ChemistryQ151–250 of 900 questions
Page 4 of 10 · English
151
ChemistryDifficultMCQMHT CET · 2024
Which of the following species acts as a base,according to the Bronsted-Lowry theory?
$HCl + NH_3 \rightleftharpoons NH_4^+{_{\text{(aq)}}} + Cl^-{_{\text{(aq)}}}$
$HCl + NH_3 \rightleftharpoons NH_4^+{_{\text{(aq)}}} + Cl^-{_{\text{(aq)}}}$
A
$Cl^-$
B
$NH_3$
C
$NH_4^+$
D
$HCl$
Solution
(B) According to the Bronsted-Lowry theory,a base is defined as a proton $(H^+)$ acceptor.
In the reaction $HCl + NH_3 \rightleftharpoons NH_4^+ + Cl^-$,$NH_3$ accepts a proton from $HCl$ to form $NH_4^+$.
Therefore,$NH_3$ acts as a Bronsted-Lowry base.
In the reaction $HCl + NH_3 \rightleftharpoons NH_4^+ + Cl^-$,$NH_3$ accepts a proton from $HCl$ to form $NH_4^+$.
Therefore,$NH_3$ acts as a Bronsted-Lowry base.
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152
ChemistryMediumMCQMHT CET · 2024
Calculate $[H_3O^{+}]$ of a monobasic acid if it is $0.04 \%$ dissociated in $0.05 \ M$ solution.
A
$1 \times 10^{-5}$
B
$1.5 \times 10^{-5}$
C
$2.0 \times 10^{-5}$
D
$3.0 \times 10^{-5}$
Solution
(C) The degree of dissociation $\alpha$ is given by $\alpha = \frac{\text{Percentage dissociation}}{100} = \frac{0.04}{100} = 4 \times 10^{-4}$.
For a monobasic acid,the concentration of hydronium ions is given by $[H_3O^{+}] = c \times \alpha$.
Substituting the values: $[H_3O^{+}] = 0.05 \ M \times 4 \times 10^{-4} = 2.0 \times 10^{-5} \ M$.
For a monobasic acid,the concentration of hydronium ions is given by $[H_3O^{+}] = c \times \alpha$.
Substituting the values: $[H_3O^{+}] = 0.05 \ M \times 4 \times 10^{-4} = 2.0 \times 10^{-5} \ M$.
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153
ChemistryMediumMCQMHT CET · 2024
Calculate $[H_3O^{+}]$ in $0.02 \ M$ solution of monobasic acid if dissociation constant is $1.8 \times 10^{-5}$.
A
$3.0 \times 10^{-4} \ M$
B
$6.0 \times 10^{-4} \ M$
C
$2.0 \times 10^{-4} \ M$
D
$4.0 \times 10^{-4} \ M$
Solution
(B) For a weak monobasic acid,the dissociation constant $K_a$ is given by $K_a = c \alpha^2$.
Thus,the degree of dissociation $\alpha = \sqrt{\frac{K_a}{c}}$.
Substituting the given values: $\alpha = \sqrt{\frac{1.8 \times 10^{-5}}{0.02}} = \sqrt{9 \times 10^{-4}} = 3 \times 10^{-2}$.
The concentration of hydronium ions is given by $[H_3O^{+}] = c \alpha$.
$[H_3O^{+}] = 0.02 \times 3 \times 10^{-2} = 6.0 \times 10^{-4} \ M$.
Thus,the degree of dissociation $\alpha = \sqrt{\frac{K_a}{c}}$.
Substituting the given values: $\alpha = \sqrt{\frac{1.8 \times 10^{-5}}{0.02}} = \sqrt{9 \times 10^{-4}} = 3 \times 10^{-2}$.
The concentration of hydronium ions is given by $[H_3O^{+}] = c \alpha$.
$[H_3O^{+}] = 0.02 \times 3 \times 10^{-2} = 6.0 \times 10^{-4} \ M$.
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154
ChemistryEasyMCQMHT CET · 2024
Calculate $\alpha$ for $0.1 \ M$ acetic acid $(K_{a} = 1.0 \times 10^{-5})$.
A
$10^{-2}$
B
$10^{-3}$
C
$10^{-4}$
D
$10^{-5}$
Solution
(A) For a weak acid $HA$,the degree of dissociation $\alpha$ is very small,so $(1 - \alpha) \cong 1$.
The formula for the degree of dissociation is $\alpha = \sqrt{\frac{K_{a}}{c}}$.
Given $K_{a} = 1.0 \times 10^{-5}$ and $c = 0.1 \ M$.
Substituting the values: $\alpha = \sqrt{\frac{1.0 \times 10^{-5}}{0.1}} = \sqrt{1.0 \times 10^{-4}} = 10^{-2}$.
The formula for the degree of dissociation is $\alpha = \sqrt{\frac{K_{a}}{c}}$.
Given $K_{a} = 1.0 \times 10^{-5}$ and $c = 0.1 \ M$.
Substituting the values: $\alpha = \sqrt{\frac{1.0 \times 10^{-5}}{0.1}} = \sqrt{1.0 \times 10^{-4}} = 10^{-2}$.
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155
ChemistryEasyMCQMHT CET · 2024
Which of the following compounds is amphoteric in nature?
A
$HCl$
B
$H_2O$
C
$CH_3COOH$
D
$NaOH$
Solution
(B) An amphoteric substance is one that can act as both an acid and a base.
$H_2O$ is a classic example of an amphoteric substance.
As an acid: $H_2O_{(l)} + NH_{3_{(aq)}} \rightleftharpoons OH_{(aq)}^{-} + NH_{4_{(aq)}}^{+}$
As a base: $H_2O_{(l)} + HCl_{(aq)} \rightleftharpoons H_3O_{(aq)}^{+} + Cl_{(aq)}^{-}$
$H_2O$ is a classic example of an amphoteric substance.
As an acid: $H_2O_{(l)} + NH_{3_{(aq)}} \rightleftharpoons OH_{(aq)}^{-} + NH_{4_{(aq)}}^{+}$
As a base: $H_2O_{(l)} + HCl_{(aq)} \rightleftharpoons H_3O_{(aq)}^{+} + Cl_{(aq)}^{-}$
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156
ChemistryMediumMCQMHT CET · 2024
Calculate the concentration of a weak monobasic acid if its degree of dissociation and dissociation constant are $5.0 \times 10^{-4}$ and $5.0 \times 10^{-9}$ respectively (in $M$)?
A
$0.1$
B
$0.02$
C
$0.03$
D
$0.04$
Solution
(B) For a weak monobasic acid,the relationship between dissociation constant $(K_{a})$,degree of dissociation $(\alpha)$,and concentration $(c)$ is given by:
$K_{a} = \alpha^2 c$
Substituting the given values:
$c = \frac{K_{a}}{\alpha^2} = \frac{5.0 \times 10^{-9}}{(5.0 \times 10^{-4})^2}$
$c = \frac{5.0 \times 10^{-9}}{25 \times 10^{-8}} = 0.2 \times 10^{-1} = 0.02 \ M$
$K_{a} = \alpha^2 c$
Substituting the given values:
$c = \frac{K_{a}}{\alpha^2} = \frac{5.0 \times 10^{-9}}{(5.0 \times 10^{-4})^2}$
$c = \frac{5.0 \times 10^{-9}}{25 \times 10^{-8}} = 0.2 \times 10^{-1} = 0.02 \ M$
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157
ChemistryMediumMCQMHT CET · 2024
Which among the following is a correct conjugate acid-base pair for the equation stated below?
$HCl + NH_3 \rightleftharpoons NH_4^{+} + Cl^{-}$
$HCl + NH_3 \rightleftharpoons NH_4^{+} + Cl^{-}$
A
$Cl^{-}$ and $NH_4^{+}$
B
$HCl$ and $NH_3$
C
$NH_4^{+}$ and $NH_3$
D
$NH_4^{+}$ and $HCl$
Solution
(C) conjugate acid-base pair differs by a single proton $(H^{+})$.
In the reaction $HCl + NH_3 \rightleftharpoons NH_4^{+} + Cl^{-}$:
$1$. $HCl$ acts as an acid and loses a proton to form its conjugate base,$Cl^{-}$. Thus,$(HCl, Cl^{-})$ is a conjugate acid-base pair.
$2$. $NH_3$ acts as a base and gains a proton to form its conjugate acid,$NH_4^{+}$. Thus,$(NH_4^{+}, NH_3)$ is a conjugate acid-base pair.
Comparing this with the given options,option $C$ represents a correct conjugate acid-base pair.
In the reaction $HCl + NH_3 \rightleftharpoons NH_4^{+} + Cl^{-}$:
$1$. $HCl$ acts as an acid and loses a proton to form its conjugate base,$Cl^{-}$. Thus,$(HCl, Cl^{-})$ is a conjugate acid-base pair.
$2$. $NH_3$ acts as a base and gains a proton to form its conjugate acid,$NH_4^{+}$. Thus,$(NH_4^{+}, NH_3)$ is a conjugate acid-base pair.
Comparing this with the given options,option $C$ represents a correct conjugate acid-base pair.
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158
ChemistryMediumMCQMHT CET · 2024
What is the pH of $10^{-8} \ M \ HCl$ solution?
A
$8$
B
$7$
C
$< 7$
D
$> 8$
Solution
(C) For a very dilute solution of $HCl$ with concentration $10^{-8} \ M$,the contribution of $H^+$ ions from the auto-ionization of water cannot be neglected.
Total $[H^+] = [H^+]_{HCl} + [H^+]_{H_2O}$.
Since $[H^+]_{HCl} = 10^{-8} \ M$ and $[H^+]_{H_2O} \approx 10^{-7} \ M$,the total $[H^+]$ will be slightly greater than $10^{-7} \ M$.
Therefore,$pH = -\log[H^+] < -\log(10^{-7}) = 7$.
Thus,the $pH$ of a $10^{-8} \ M \ HCl$ solution is slightly less than $7$.
Total $[H^+] = [H^+]_{HCl} + [H^+]_{H_2O}$.
Since $[H^+]_{HCl} = 10^{-8} \ M$ and $[H^+]_{H_2O} \approx 10^{-7} \ M$,the total $[H^+]$ will be slightly greater than $10^{-7} \ M$.
Therefore,$pH = -\log[H^+] < -\log(10^{-7}) = 7$.
Thus,the $pH$ of a $10^{-8} \ M \ HCl$ solution is slightly less than $7$.
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159
ChemistryDifficultMCQMHT CET · 2024
Calculate the dissociation constant of a weak monobasic acid if it is $0.05 \%$ dissociated in a $0.02 \ M$ solution.
A
$2.0 \times 10^{-9}$
B
$3.0 \times 10^{-9}$
C
$4.0 \times 10^{-9}$
D
$5.0 \times 10^{-9}$
Solution
(D) For a weak monobasic acid $HA$,the dissociation is given by: $HA \rightleftharpoons H^+_{(aq)} + A^-_{(aq)}$
Given degree of dissociation $\alpha = 0.05 \% = 0.05 \times 10^{-2} = 5 \times 10^{-4}$.
Concentration $C = 0.02 \ M = 2 \times 10^{-2} \ M$.
For a weak acid,the dissociation constant $K_a$ is given by the formula $K_a = \alpha^2 C$ (since $\alpha \ll 1$).
Substituting the values: $K_a = (5 \times 10^{-4})^2 \times (2 \times 10^{-2})$.
$K_a = (25 \times 10^{-8}) \times (2 \times 10^{-2}) = 50 \times 10^{-10} = 5.0 \times 10^{-9}$.
Given degree of dissociation $\alpha = 0.05 \% = 0.05 \times 10^{-2} = 5 \times 10^{-4}$.
Concentration $C = 0.02 \ M = 2 \times 10^{-2} \ M$.
For a weak acid,the dissociation constant $K_a$ is given by the formula $K_a = \alpha^2 C$ (since $\alpha \ll 1$).
Substituting the values: $K_a = (5 \times 10^{-4})^2 \times (2 \times 10^{-2})$.
$K_a = (25 \times 10^{-8}) \times (2 \times 10^{-2}) = 50 \times 10^{-10} = 5.0 \times 10^{-9}$.
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160
ChemistryMediumMCQMHT CET · 2024
Which among the following salts turns red litmus blue in its aqueous solution?
A
$Na_2SO_4$
B
$CH_3COONa$
C
$NH_4NO_3$
D
$CuCl_2$
Solution
(B) The $pH$ of an aqueous salt solution depends on the nature of the acid and base from which it is derived.
$Na_2SO_4$ is a salt of a strong acid $(H_2SO_4)$ and a strong base $(NaOH)$,making it neutral.
$NH_4NO_3$ and $CuCl_2$ are salts of a weak base and a strong acid,making them acidic.
$CH_3COONa$ is a salt of a weak acid $(CH_3COOH)$ and a strong base $(NaOH)$.
In water,the acetate ion $(CH_3COO^-)$ undergoes hydrolysis to produce $OH^-$ ions,making the solution basic.
Basic solutions turn red litmus blue.
$Na_2SO_4$ is a salt of a strong acid $(H_2SO_4)$ and a strong base $(NaOH)$,making it neutral.
$NH_4NO_3$ and $CuCl_2$ are salts of a weak base and a strong acid,making them acidic.
$CH_3COONa$ is a salt of a weak acid $(CH_3COOH)$ and a strong base $(NaOH)$.
In water,the acetate ion $(CH_3COO^-)$ undergoes hydrolysis to produce $OH^-$ ions,making the solution basic.
Basic solutions turn red litmus blue.
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161
ChemistryEasyMCQMHT CET · 2024
$A$ monobasic acid is $5 \%$ dissociated in its $0.02 \ M$ solution. Calculate the dissociation constant of the acid.
A
$2 \times 10^{-2}$
B
$4 \times 10^{-4}$
C
$5 \times 10^{-5}$
D
$2.5 \times 10^{-4}$
Solution
(C) The degree of dissociation $\alpha$ is given by $\alpha = \frac{\text{Percent dissociation}}{100} = \frac{5}{100} = 0.05 = 5 \times 10^{-2}$.
Given concentration $c = 0.02 \ M = 2 \times 10^{-2} \ M$.
For a weak monobasic acid,the dissociation constant $K_a$ is given by the formula $K_a = \alpha^2 c$ (since $\alpha$ is very small,$1 - \alpha \approx 1$).
Substituting the values: $K_a = (5 \times 10^{-2})^2 \times (2 \times 10^{-2})$.
$K_a = (25 \times 10^{-4}) \times (2 \times 10^{-2}) = 50 \times 10^{-6} = 5 \times 10^{-5}$.
Given concentration $c = 0.02 \ M = 2 \times 10^{-2} \ M$.
For a weak monobasic acid,the dissociation constant $K_a$ is given by the formula $K_a = \alpha^2 c$ (since $\alpha$ is very small,$1 - \alpha \approx 1$).
Substituting the values: $K_a = (5 \times 10^{-2})^2 \times (2 \times 10^{-2})$.
$K_a = (25 \times 10^{-4}) \times (2 \times 10^{-2}) = 50 \times 10^{-6} = 5 \times 10^{-5}$.
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162
ChemistryEasyMCQMHT CET · 2024
Which among the following is the conjugate base of $HClO_4$?
A
$ClO_4^{2-}$
B
$ClO_4$
C
$HCl$
D
$ClO_4^{-}$
Solution
(D) conjugate base is formed when a Bronsted-Lowry acid donates a proton $(H^{+})$.
For the acid $HClO_4$,the reaction is:
$HClO_4 \rightarrow H^{+} + ClO_4^{-}$
Therefore,the conjugate base of $HClO_4$ is $ClO_4^{-}$.
For the acid $HClO_4$,the reaction is:
$HClO_4 \rightarrow H^{+} + ClO_4^{-}$
Therefore,the conjugate base of $HClO_4$ is $ClO_4^{-}$.
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163
ChemistryEasyMCQMHT CET · 2024
Conjugate acid of $NH_2^{-}$ and $NH_3$ are respectively
A
$NH_4OH$ and $NH_2OH$
B
$NH_3$ and $NH_2^{-}$
C
$NH_3$ and $NH_4^{+}$
D
$NH_4^{+}$ and $NH_3$
Solution
(C) conjugate acid is formed when a Bronsted-Lowry base accepts a proton $(H^{+})$.
For $NH_2^{-}$,the conjugate acid is $NH_2^{-} + H^{+} \rightarrow NH_3$.
For $NH_3$,the conjugate acid is $NH_3 + H^{+} \rightarrow NH_4^{+}$.
Therefore,the conjugate acids are $NH_3$ and $NH_4^{+}$ respectively.
For $NH_2^{-}$,the conjugate acid is $NH_2^{-} + H^{+} \rightarrow NH_3$.
For $NH_3$,the conjugate acid is $NH_3 + H^{+} \rightarrow NH_4^{+}$.
Therefore,the conjugate acids are $NH_3$ and $NH_4^{+}$ respectively.
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164
ChemistryEasyMCQMHT CET · 2024
Calculate the $pH$ of $0.02 \ M$ monobasic acid having $2 \%$ dissociation.
A
$3.4$
B
$4.5$
C
$5.1$
D
$5.8$
Solution
(A) For a monobasic acid $HA$,the dissociation is given by: $HA \rightleftharpoons H_{(aq)}^{+} + A_{(aq)}^{-}$
Concentration of $H^{+}$ ions is calculated as $[H^{+}] = \alpha \times C$,where $\alpha$ is the degree of dissociation and $C$ is the molar concentration.
Given $\alpha = 2 \% = 0.02$ and $C = 0.02 \ M$.
$[H^{+}] = 0.02 \times 0.02 = 0.0004 \ M = 4 \times 10^{-4} \ M$.
$pH = -\log[H^{+}] = -\log(4 \times 10^{-4}) = 4 - \log(4) = 4 - 0.602 = 3.398 \approx 3.4$.
Concentration of $H^{+}$ ions is calculated as $[H^{+}] = \alpha \times C$,where $\alpha$ is the degree of dissociation and $C$ is the molar concentration.
Given $\alpha = 2 \% = 0.02$ and $C = 0.02 \ M$.
$[H^{+}] = 0.02 \times 0.02 = 0.0004 \ M = 4 \times 10^{-4} \ M$.
$pH = -\log[H^{+}] = -\log(4 \times 10^{-4}) = 4 - \log(4) = 4 - 0.602 = 3.398 \approx 3.4$.
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165
ChemistryMediumMCQMHT CET · 2024
Which of the following substances acts as a base when reacted with water?
A
$CH_3COOH$
B
$H_2C_2O_4$
C
$HCl$
D
$NH_3$
Solution
(D) When $NH_3$ reacts with water,it accepts a proton $(H^+)$ from water to form $NH_4^+$ and $OH^-$ ions.
Since it accepts a proton,it acts as a Brønsted-Lowry base.
The reaction is: $NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-$
Here,$NH_3$ acts as a base and $H_2O$ acts as an acid.
Since it accepts a proton,it acts as a Brønsted-Lowry base.
The reaction is: $NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-$
Here,$NH_3$ acts as a base and $H_2O$ acts as an acid.
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166
ChemistryEasyMCQMHT CET · 2024
Calculate the $pH$ of a $0.002 \ M$ $KOH$ solution.
A
$10.4$
B
$11.3$
C
$12.4$
D
$13.2$
Solution
(B) $KOH$ is a strong base,so it dissociates completely: $[OH^-] = [KOH] = 0.002 \ M$.
$pOH = -\log_{10}[OH^-] = -\log_{10}(2 \times 10^{-3}) = 3 - \log_{10}(2) = 3 - 0.301 = 2.699 \approx 2.7$.
Using the relation $pH + pOH = 14$ at $298 \ K$:
$pH = 14 - 2.7 = 11.3$.
$pOH = -\log_{10}[OH^-] = -\log_{10}(2 \times 10^{-3}) = 3 - \log_{10}(2) = 3 - 0.301 = 2.699 \approx 2.7$.
Using the relation $pH + pOH = 14$ at $298 \ K$:
$pH = 14 - 2.7 = 11.3$.
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167
ChemistryDifficultMCQMHT CET · 2024
Calculate the $[OH^{-}]$ if $pOH$ of a solution is $4.94$.
A
$2.356 \times 10^{-5} \ M$
B
$1.881 \times 10^{-5} \ M$
C
$1.417 \times 10^{-5} \ M$
D
$1.148 \times 10^{-5} \ M$
Solution
(D) Given: $pOH = 4.94$.
We know that,$pOH = -\log[OH^{-}]$.
Therefore,$[OH^{-}] = 10^{-pOH}$.
$[OH^{-}] = 10^{-4.94}$.
To solve this,we can write $10^{-4.94}$ as $10^{0.06 - 5} = 10^{0.06} \times 10^{-5}$.
Since $\text{antilog}(0.06) \approx 1.148$,we get $[OH^{-}] = 1.148 \times 10^{-5} \ M$.
Thus,the correct option is $(d)$.
We know that,$pOH = -\log[OH^{-}]$.
Therefore,$[OH^{-}] = 10^{-pOH}$.
$[OH^{-}] = 10^{-4.94}$.
To solve this,we can write $10^{-4.94}$ as $10^{0.06 - 5} = 10^{0.06} \times 10^{-5}$.
Since $\text{antilog}(0.06) \approx 1.148$,we get $[OH^{-}] = 1.148 \times 10^{-5} \ M$.
Thus,the correct option is $(d)$.
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168
ChemistryMediumMCQMHT CET · 2024
Dissociation constant and degree of dissociation of a weak acid are $1.8 \times 10^{-5}$ and $0.03$ respectively. What will be the concentration of the solution of the weak acid (in $M$)?
A
$0.2$
B
$0.02$
C
$0.5$
D
$0.05$
Solution
(B) For a weak acid,the relationship between dissociation constant $(K_a)$,degree of dissociation $(\alpha)$,and concentration $(c)$ is given by $K_a = c \alpha^2$.
Given: $K_a = 1.8 \times 10^{-5}$ and $\alpha = 0.03$.
Rearranging the formula to solve for concentration: $c = \frac{K_a}{\alpha^2}$.
Substituting the values: $c = \frac{1.8 \times 10^{-5}}{(0.03)^2} = \frac{1.8 \times 10^{-5}}{9 \times 10^{-4}}$.
Calculating the result: $c = 0.2 \times 10^{-1} \ M = 0.02 \ M$.
Given: $K_a = 1.8 \times 10^{-5}$ and $\alpha = 0.03$.
Rearranging the formula to solve for concentration: $c = \frac{K_a}{\alpha^2}$.
Substituting the values: $c = \frac{1.8 \times 10^{-5}}{(0.03)^2} = \frac{1.8 \times 10^{-5}}{9 \times 10^{-4}}$.
Calculating the result: $c = 0.2 \times 10^{-1} \ M = 0.02 \ M$.
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169
ChemistryDifficultMCQMHT CET · 2024
What is the $pOH$ of a $1 \ mM$ solution of $Ca(OH)_2$?
A
$2.7$
B
$10.3$
C
$12.3$
D
$11.3$
Solution
(A) The concentration of the solution is $1 \ mM = 10^{-3} \ M$.
Since $Ca(OH)_2$ is a strong base,it dissociates completely as: $Ca(OH)_2 \rightarrow Ca^{2+} + 2OH^-$.
For every $1 \text{ mole}$ of $Ca(OH)_2$,$2 \text{ moles}$ of $OH^-$ ions are produced.
Therefore,$[OH^-] = 2 \times 10^{-3} \ M$.
$pOH = -\log_{10} [OH^-] = -\log_{10} (2 \times 10^{-3})$.
$pOH = -(\log_{10} 2 + \log_{10} 10^{-3}) = -(0.301 - 3) = 2.699 \approx 2.7$.
Since $Ca(OH)_2$ is a strong base,it dissociates completely as: $Ca(OH)_2 \rightarrow Ca^{2+} + 2OH^-$.
For every $1 \text{ mole}$ of $Ca(OH)_2$,$2 \text{ moles}$ of $OH^-$ ions are produced.
Therefore,$[OH^-] = 2 \times 10^{-3} \ M$.
$pOH = -\log_{10} [OH^-] = -\log_{10} (2 \times 10^{-3})$.
$pOH = -(\log_{10} 2 + \log_{10} 10^{-3}) = -(0.301 - 3) = 2.699 \approx 2.7$.
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170
ChemistryMediumMCQMHT CET · 2024
What is the $pH$ of a centimolar solution of $H_2SO_4$?
A
$1.7$
B
$2$
C
$3.2$
D
$4.5$
Solution
(A) $H_2SO_4$ is a strong diprotic acid,so it dissociates completely as $H_2SO_4 \rightarrow 2H^+ + SO_4^{2-}$.
For a centimolar solution,concentration $C = 0.01 \ M = 10^{-2} \ M$.
The concentration of $H^+$ ions is $[H^+] = 2 \times C = 2 \times 10^{-2} \ M$.
$pH = -\log_{10} [H^+] = -\log_{10} (2 \times 10^{-2})$.
$pH = -(\log_{10} 2 + \log_{10} 10^{-2}) = -(\log_{10} 2 - 2) = 2 - \log_{10} 2$.
Using $\log_{10} 2 \approx 0.3010$,we get $pH = 2 - 0.3010 = 1.699 \approx 1.7$.
For a centimolar solution,concentration $C = 0.01 \ M = 10^{-2} \ M$.
The concentration of $H^+$ ions is $[H^+] = 2 \times C = 2 \times 10^{-2} \ M$.
$pH = -\log_{10} [H^+] = -\log_{10} (2 \times 10^{-2})$.
$pH = -(\log_{10} 2 + \log_{10} 10^{-2}) = -(\log_{10} 2 - 2) = 2 - \log_{10} 2$.
Using $\log_{10} 2 \approx 0.3010$,we get $pH = 2 - 0.3010 = 1.699 \approx 1.7$.
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171
ChemistryMediumMCQMHT CET · 2024
The dissociation constant of a weak monobasic acid is $3.2 \times 10^{-4}$. Calculate the degree of dissociation in its $0.04 \ M$ solution.
A
$0.0128$
B
$0.0151$
C
$0.078$
D
$0.089$
Solution
(D) Given: Dissociation constant $K_a = 3.2 \times 10^{-4}$ and concentration $c = 0.04 \ M$.
For a weak monobasic acid,the degree of dissociation $\alpha$ is given by the formula:
$\alpha = \sqrt{\frac{K_a}{c}}$
Substituting the values:
$\alpha = \sqrt{\frac{3.2 \times 10^{-4}}{0.04}} = \sqrt{\frac{320 \times 10^{-6}}{4 \times 10^{-2}}} = \sqrt{80 \times 10^{-4}} = \sqrt{8 \times 10^{-3}} \approx 0.0894$
Thus,the degree of dissociation is approximately $0.089$.
For a weak monobasic acid,the degree of dissociation $\alpha$ is given by the formula:
$\alpha = \sqrt{\frac{K_a}{c}}$
Substituting the values:
$\alpha = \sqrt{\frac{3.2 \times 10^{-4}}{0.04}} = \sqrt{\frac{320 \times 10^{-6}}{4 \times 10^{-2}}} = \sqrt{80 \times 10^{-4}} = \sqrt{8 \times 10^{-3}} \approx 0.0894$
Thus,the degree of dissociation is approximately $0.089$.
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172
ChemistryDifficultMCQMHT CET · 2024
$A$ buffer solution is prepared by mixing $0.2 \ M \ NH_4OH$ and $1 \ M \ NH_4Cl$. What is the $pH$ value of the buffer solution? (Given $pK_b = 4.744$)
A
$9.256$
B
$8.556$
C
$5.444$
D
$4.744$
Solution
(B) For a basic buffer solution,the $pOH$ is calculated using the Henderson-Hasselbalch equation:
$pOH = pK_b + \log_{10} \frac{[Salt]}{[Base]}$
Given $pK_b = 4.744$,$[Salt] = [NH_4Cl] = 1 \ M$,and $[Base] = [NH_4OH] = 0.2 \ M$.
$pOH = 4.744 + \log_{10} \frac{1}{0.2} = 4.744 + \log_{10} (5)$
$pOH = 4.744 + 0.699 = 5.443$
Since $pH + pOH = 14$ at $25^{\circ}C$,
$pH = 14 - 5.443 = 8.557 \approx 8.56$.
$pOH = pK_b + \log_{10} \frac{[Salt]}{[Base]}$
Given $pK_b = 4.744$,$[Salt] = [NH_4Cl] = 1 \ M$,and $[Base] = [NH_4OH] = 0.2 \ M$.
$pOH = 4.744 + \log_{10} \frac{1}{0.2} = 4.744 + \log_{10} (5)$
$pOH = 4.744 + 0.699 = 5.443$
Since $pH + pOH = 14$ at $25^{\circ}C$,
$pH = 14 - 5.443 = 8.557 \approx 8.56$.
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173
ChemistryDifficultMCQMHT CET · 2024
An aqueous solution of a strong monoacidic base has a concentration of $1 \times 10^{-4} \ M$. What is the value of $pH$ at $25^{\circ} C$?
A
$7$
B
$4$
C
$3$
D
$10$
Solution
(D) For a strong monoacidic base,the concentration of hydroxide ions is $[OH^-] = 1 \times 10^{-4} \ M$.
$pOH = -\log_{10}[OH^-]$
$pOH = -\log_{10}(1 \times 10^{-4}) = 4$
Since $pH + pOH = 14$ at $25^{\circ} C$,
$pH = 14 - pOH = 14 - 4 = 10$.
$pOH = -\log_{10}[OH^-]$
$pOH = -\log_{10}(1 \times 10^{-4}) = 4$
Since $pH + pOH = 14$ at $25^{\circ} C$,
$pH = 14 - pOH = 14 - 4 = 10$.
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174
ChemistryDifficultMCQMHT CET · 2024
Calculate the $pH$ of $0.01 \ M$ sulphuric acid.
A
$1.699$
B
$2$
C
$0.699$
D
$3.398$
Solution
(A) $H_2SO_4$ is a strong diprotic acid that dissociates completely in water:
$H_2SO_{4(aq)} + 2H_2O_{(l)} \longrightarrow 2H_3O^{+}_{(aq)} + SO_{4(aq)}^{2-}$
Since it is a strong acid,the concentration of hydronium ions is twice the concentration of the acid:
$[H_3O^{+}] = 2 \times c = 2 \times 0.01 \ M = 0.02 \ M = 2 \times 10^{-2} \ M$
The $pH$ is calculated as:
$pH = -\log_{10}[H_3O^{+}]$
$pH = -\log_{10}(2 \times 10^{-2})$
$pH = -(\log_{10}2 + \log_{10}10^{-2})$
$pH = -(0.3010 - 2)$
$pH = 2 - 0.3010 = 1.699$
$H_2SO_{4(aq)} + 2H_2O_{(l)} \longrightarrow 2H_3O^{+}_{(aq)} + SO_{4(aq)}^{2-}$
Since it is a strong acid,the concentration of hydronium ions is twice the concentration of the acid:
$[H_3O^{+}] = 2 \times c = 2 \times 0.01 \ M = 0.02 \ M = 2 \times 10^{-2} \ M$
The $pH$ is calculated as:
$pH = -\log_{10}[H_3O^{+}]$
$pH = -\log_{10}(2 \times 10^{-2})$
$pH = -(\log_{10}2 + \log_{10}10^{-2})$
$pH = -(0.3010 - 2)$
$pH = 2 - 0.3010 = 1.699$
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175
ChemistryEasyMCQMHT CET · 2024
Which of the following salts is $NOT$ derived from a weak acid and a weak base?
A
Ammonium fluoride
B
Ammonium cyanide
C
Ammonium acetate
D
Ammonium chloride
Solution
(D) salt derived from a weak acid and a weak base is formed by the reaction of a weak acid and a weak base.
$NH_4F$ is formed from $NH_4OH$ (weak base) and $HF$ (weak acid).
$NH_4CN$ is formed from $NH_4OH$ (weak base) and $HCN$ (weak acid).
$CH_3COONH_4$ is formed from $NH_4OH$ (weak base) and $CH_3COOH$ (weak acid).
$NH_4Cl$ is formed from $NH_4OH$ (weak base) and $HCl$ (strong acid).
Therefore,$NH_4Cl$ is $NOT$ derived from a weak acid and a weak base.
$NH_4F$ is formed from $NH_4OH$ (weak base) and $HF$ (weak acid).
$NH_4CN$ is formed from $NH_4OH$ (weak base) and $HCN$ (weak acid).
$CH_3COONH_4$ is formed from $NH_4OH$ (weak base) and $CH_3COOH$ (weak acid).
$NH_4Cl$ is formed from $NH_4OH$ (weak base) and $HCl$ (strong acid).
Therefore,$NH_4Cl$ is $NOT$ derived from a weak acid and a weak base.
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176
ChemistryMediumMCQMHT CET · 2024
What is the value of $pOH$ if a buffer solution is prepared by mixing equal volumes of $0.4 \ M$ $NH_4OH$ and $0.5 \ M$ $NH_4Cl$ solutions? $(pK_b = 4.730)$
A
$6$
B
$4.83$
C
$10.42$
D
$7.81$
Solution
(B) For a basic buffer,the Henderson-Hasselbalch equation is given by:
$pOH = pK_b + \log \frac{[Salt]}{[Base]}$
Since equal volumes are mixed,the new concentrations are halved,but the ratio of $[Salt]/[Base]$ remains the same as the ratio of their initial concentrations.
$[Salt] = [NH_4Cl] = 0.5 \ M$
$[Base] = [NH_4OH] = 0.4 \ M$
$pOH = 4.730 + \log \left( \frac{0.5}{0.4} \right)$
$pOH = 4.730 + \log(1.25)$
$pOH = 4.730 + 0.0969 \approx 4.8269 \approx 4.83$
$pOH = pK_b + \log \frac{[Salt]}{[Base]}$
Since equal volumes are mixed,the new concentrations are halved,but the ratio of $[Salt]/[Base]$ remains the same as the ratio of their initial concentrations.
$[Salt] = [NH_4Cl] = 0.5 \ M$
$[Base] = [NH_4OH] = 0.4 \ M$
$pOH = 4.730 + \log \left( \frac{0.5}{0.4} \right)$
$pOH = 4.730 + \log(1.25)$
$pOH = 4.730 + 0.0969 \approx 4.8269 \approx 4.83$
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177
ChemistryDifficultMCQMHT CET · 2024
What is the ratio of concentration of salt to concentration of weak acid in a buffer solution to maintain its $pH$ value at $7.2$ $(pK_{a} = 6.2)$?
A
$1.5$
B
$10$
C
$5$
D
$8.5$
Solution
(B) The Henderson-Hasselbalch equation for an acidic buffer is given by: $pH = pK_{a} + \log \frac{[Salt]}{[Acid]}$
Given $pH = 7.2$ and $pK_{a} = 6.2$.
Substituting the values: $7.2 = 6.2 + \log \frac{[Salt]}{[Acid]}$
$7.2 - 6.2 = \log \frac{[Salt]}{[Acid]}$
$1 = \log \frac{[Salt]}{[Acid]}$
Since $\log_{10} 10 = 1$,we have $\frac{[Salt]}{[Acid]} = 10$.
Given $pH = 7.2$ and $pK_{a} = 6.2$.
Substituting the values: $7.2 = 6.2 + \log \frac{[Salt]}{[Acid]}$
$7.2 - 6.2 = \log \frac{[Salt]}{[Acid]}$
$1 = \log \frac{[Salt]}{[Acid]}$
Since $\log_{10} 10 = 1$,we have $\frac{[Salt]}{[Acid]} = 10$.
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178
ChemistryDifficultMCQMHT CET · 2024
Calculate the $pH$ of a buffer solution containing $0.027 \ M$ weak acid and $0.054 \ M$ of its salt with a strong base,if the $pK_{a}$ is $4.2$.
A
$4.5$
B
$3.2$
C
$5.6$
D
$6.4$
Solution
(A) For an acidic buffer solution,the Henderson-Hasselbalch equation is given by:
$pH = pK_{a} + \log_{10} \frac{[\text{Salt}]}{[\text{Acid}]}$
Given:
$pK_{a} = 4.2$
$[\text{Salt}] = 0.054 \ M$
$[\text{Acid}] = 0.027 \ M$
Substituting the values:
$pH = 4.2 + \log_{10} \left( \frac{0.054}{0.027} \right)$
$pH = 4.2 + \log_{10} (2)$
Since $\log_{10} 2 \approx 0.3010$,
$pH = 4.2 + 0.3010 = 4.5010$
Thus,the $pH$ is approximately $4.5$.
$pH = pK_{a} + \log_{10} \frac{[\text{Salt}]}{[\text{Acid}]}$
Given:
$pK_{a} = 4.2$
$[\text{Salt}] = 0.054 \ M$
$[\text{Acid}] = 0.027 \ M$
Substituting the values:
$pH = 4.2 + \log_{10} \left( \frac{0.054}{0.027} \right)$
$pH = 4.2 + \log_{10} (2)$
Since $\log_{10} 2 \approx 0.3010$,
$pH = 4.2 + 0.3010 = 4.5010$
Thus,the $pH$ is approximately $4.5$.
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179
ChemistryMediumMCQMHT CET · 2024
Calculate the $pH$ of a buffer solution containing $0.35 \ M$ weak acid and $0.70 \ M$ of its salt with a strong base if $pK_{a}$ is $4.56$.
A
$6.11$
B
$3.72$
C
$4.86$
D
$5.65$
Solution
(C) For an acidic buffer,the Henderson-Hasselbalch equation is given by:
$pH = pK_{a} + \log_{10} \frac{[\text{salt}]}{[\text{acid}]}$
Given:
$pK_{a} = 4.56$
$[\text{salt}] = 0.70 \ M$
$[\text{acid}] = 0.35 \ M$
Substituting the values:
$pH = 4.56 + \log_{10} \frac{0.70}{0.35}$
$pH = 4.56 + \log_{10} (2)$
Since $\log_{10} (2) \approx 0.3010$:
$pH = 4.56 + 0.3010 = 4.861$
Thus,the $pH$ of the buffer solution is $4.86$.
$pH = pK_{a} + \log_{10} \frac{[\text{salt}]}{[\text{acid}]}$
Given:
$pK_{a} = 4.56$
$[\text{salt}] = 0.70 \ M$
$[\text{acid}] = 0.35 \ M$
Substituting the values:
$pH = 4.56 + \log_{10} \frac{0.70}{0.35}$
$pH = 4.56 + \log_{10} (2)$
Since $\log_{10} (2) \approx 0.3010$:
$pH = 4.56 + 0.3010 = 4.861$
Thus,the $pH$ of the buffer solution is $4.86$.
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180
ChemistryEasyMCQMHT CET · 2024
Which of the following mixtures in water acts as a basic buffer?
A
$NH_4OH + NH_4Cl$
B
$C_6H_5COOH + C_6H_5COONa$
C
$HCOOH + HCOOK$
D
$CH_3COOH + CH_3COONa$
Solution
(A) basic buffer is a mixture of a weak base and its salt with a strong acid.
In option $A$,$NH_4OH$ is a weak base and $NH_4Cl$ is a salt of a strong acid $(HCl)$ and a weak base $(NH_4OH)$.
Therefore,$NH_4OH + NH_4Cl$ acts as a basic buffer.
Options $B$,$C$,and $D$ consist of a weak acid and its salt with a strong base,which are examples of acidic buffers.
In option $A$,$NH_4OH$ is a weak base and $NH_4Cl$ is a salt of a strong acid $(HCl)$ and a weak base $(NH_4OH)$.
Therefore,$NH_4OH + NH_4Cl$ acts as a basic buffer.
Options $B$,$C$,and $D$ consist of a weak acid and its salt with a strong base,which are examples of acidic buffers.
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181
ChemistryEasyMCQMHT CET · 2024
Calculate the $pH$ of a buffer solution containing $0.04 \ M$ $NaF$ and $0.02 \ M$ $HF$ $[pK_a = 3.142]$.
A
$4.5$
B
$3.4$
C
$2.6$
D
$5.1$
Solution
(B) The solution containing $0.04 \ M$ $NaF$ (salt) and $0.02 \ M$ $HF$ (weak acid) forms an acidic buffer.
Using the Henderson-Hasselbalch equation:
$pH = pK_a + \log_{10} \frac{[Salt]}{[Acid]}$
Substituting the given values:
$pH = 3.142 + \log_{10} \frac{0.04}{0.02}$
$pH = 3.142 + \log_{10}(2)$
Since $\log_{10}(2) \approx 0.3010$:
$pH = 3.142 + 0.3010 = 3.443$
Rounding to one decimal place,we get $pH \approx 3.4$.
Using the Henderson-Hasselbalch equation:
$pH = pK_a + \log_{10} \frac{[Salt]}{[Acid]}$
Substituting the given values:
$pH = 3.142 + \log_{10} \frac{0.04}{0.02}$
$pH = 3.142 + \log_{10}(2)$
Since $\log_{10}(2) \approx 0.3010$:
$pH = 3.142 + 0.3010 = 3.443$
Rounding to one decimal place,we get $pH \approx 3.4$.
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182
ChemistryMediumMCQMHT CET · 2024
Which of the following buffers is used to maintain the $pH$ of human blood naturally?
A
Hydrogen cyanide and sodium cyanide
B
Copper hydroxide and copper chloride
C
Carbonic acid and salt of carbonic acid
D
Ammonium hydroxide and ammonium chloride
Solution
(C) The $pH$ of human blood is maintained naturally at $7.36-7.42$ by the $(HCO_3^{-} + H_2CO_3)$ buffer system.
This buffer system consists of carbonic acid $(H_2CO_3)$ and its conjugate base,the bicarbonate ion $(HCO_3^{-})$.
This buffer system consists of carbonic acid $(H_2CO_3)$ and its conjugate base,the bicarbonate ion $(HCO_3^{-})$.
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183
ChemistryMediumMCQMHT CET · 2024
$A$ buffer solution contains equal concentrations of weak acid and its salt with a strong base. Calculate the $pH$ of the buffer solution if the dissociation constant of the weak acid is $1.8 \times 10^{-5}$.
A
$4.7447$
B
$5.142$
C
$5.8496$
D
$4.0128$
Solution
(A) For an acidic buffer,the Henderson-Hasselbalch equation is given by:
$pH = pK_{a} + \log \frac{[\text{Salt}]}{[\text{Acid}]}$
Given that the concentrations of the weak acid and its salt are equal,i.e.,$[\text{Salt}] = [\text{Acid}]$.
Therefore,the equation simplifies to:
$pH = pK_{a} = -\log K_{a}$
Given $K_{a} = 1.8 \times 10^{-5}$.
$pH = -\log (1.8 \times 10^{-5}) = -(\log 1.8 + \log 10^{-5})$
$pH = -(\log 1.8 - 5) = 5 - \log 1.8$
Using $\log 1.8 \approx 0.2553$:
$pH = 5 - 0.2553 = 4.7447$
$pH = pK_{a} + \log \frac{[\text{Salt}]}{[\text{Acid}]}$
Given that the concentrations of the weak acid and its salt are equal,i.e.,$[\text{Salt}] = [\text{Acid}]$.
Therefore,the equation simplifies to:
$pH = pK_{a} = -\log K_{a}$
Given $K_{a} = 1.8 \times 10^{-5}$.
$pH = -\log (1.8 \times 10^{-5}) = -(\log 1.8 + \log 10^{-5})$
$pH = -(\log 1.8 - 5) = 5 - \log 1.8$
Using $\log 1.8 \approx 0.2553$:
$pH = 5 - 0.2553 = 4.7447$
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184
ChemistryDifficultMCQMHT CET · 2024
$A$ buffer solution is prepared by mixing $0.01 \ M$ $HCN$ and $0.02 \ M$ $NaCN$. If $K_{a}$ for $HCN$ is $6.6 \times 10^{-10}$,what is the concentration of $H^{+}$ ions in the solution?
A
$3.3 \times 10^{-6} \ M$
B
$3.3 \times 10^{-10} \ M$
C
$1.32 \times 10^{-6} \ M$
D
$1.32 \times 10^{-10} \ M$
Solution
(B) The concentration of $H^{+}$ ions in an acidic buffer can be calculated using the Henderson-Hasselbalch equation:
$[H^{+}] = K_{a} \times \frac{[Acid]}{[Salt]}$
Given:
$K_{a} = 6.6 \times 10^{-10}$
$[Acid] = [HCN] = 0.01 \ M$
$[Salt] = [NaCN] = 0.02 \ M$
Substituting the values:
$[H^{+}] = (6.6 \times 10^{-10}) \times \frac{0.01}{0.02}$
$[H^{+}] = (6.6 \times 10^{-10}) \times 0.5$
$[H^{+}] = 3.3 \times 10^{-10} \ M$
$[H^{+}] = K_{a} \times \frac{[Acid]}{[Salt]}$
Given:
$K_{a} = 6.6 \times 10^{-10}$
$[Acid] = [HCN] = 0.01 \ M$
$[Salt] = [NaCN] = 0.02 \ M$
Substituting the values:
$[H^{+}] = (6.6 \times 10^{-10}) \times \frac{0.01}{0.02}$
$[H^{+}] = (6.6 \times 10^{-10}) \times 0.5$
$[H^{+}] = 3.3 \times 10^{-10} \ M$
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185
ChemistryMediumMCQMHT CET · 2024
An acidic buffer solution is prepared by mixing a proportionate quantity of
A
strong acid and its salt with a weak base.
B
strong base and its salt with a weak acid.
C
weak acid and its salt with a strong base.
D
weak base and its salt with a strong acid.
Solution
(C) An acidic buffer is prepared by mixing a weak acid and its salt with a strong base.
For example,a mixture of $CH_3COOH$ (weak acid) and $CH_3COONa$ (salt of weak acid and strong base) acts as an acidic buffer.
The weak acid and its conjugate base (provided by the salt) work together to resist changes in $pH$ upon the addition of small amounts of acid or base.
For example,a mixture of $CH_3COOH$ (weak acid) and $CH_3COONa$ (salt of weak acid and strong base) acts as an acidic buffer.
The weak acid and its conjugate base (provided by the salt) work together to resist changes in $pH$ upon the addition of small amounts of acid or base.
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186
ChemistryMediumMCQMHT CET · 2024
Which of the following mixtures in water acts as a buffer?
A
Acetic acid and sodium acetate
B
Acetic acid and ammonium chloride
C
Ammonium hydroxide and sodium chloride
D
Formic acid and acetic acid
Solution
(A) buffer solution is formed by a mixture of a weak acid and its conjugate base (salt of the weak acid with a strong base).
$CH_3COOH$ is a weak acid and $CH_3COONa$ is its salt with a strong base $(NaOH)$.
Therefore,the mixture of $CH_3COOH$ and $CH_3COONa$ acts as an acidic buffer solution in water.
$CH_3COOH$ is a weak acid and $CH_3COONa$ is its salt with a strong base $(NaOH)$.
Therefore,the mixture of $CH_3COOH$ and $CH_3COONa$ acts as an acidic buffer solution in water.
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187
ChemistryMediumMCQMHT CET · 2024
Calculate the solubility product of sparingly soluble salt $BA$ at $300 \ K$ if its solubility is $9.1 \times 10^{-3} \ mol \ dm^{-3}$ at same temperature.
A
$9.635 \times 10^{-5}$
B
$9.012 \times 10^{-5}$
C
$8.281 \times 10^{-5}$
D
$7.816 \times 10^{-5}$
Solution
(C) The dissociation of the salt $BA$ is given by: $BA_{(s)} \rightleftharpoons B^{+}_{(aq)} + A^{-}_{(aq)}$
For a salt of type $BA$,the solubility product $K_{sp}$ is related to solubility $S$ by the formula: $K_{sp} = S^2$
Given solubility $S = 9.1 \times 10^{-3} \ mol \ dm^{-3}$
$K_{sp} = (9.1 \times 10^{-3})^2$
$K_{sp} = 82.81 \times 10^{-6} = 8.281 \times 10^{-5}$
For a salt of type $BA$,the solubility product $K_{sp}$ is related to solubility $S$ by the formula: $K_{sp} = S^2$
Given solubility $S = 9.1 \times 10^{-3} \ mol \ dm^{-3}$
$K_{sp} = (9.1 \times 10^{-3})^2$
$K_{sp} = 82.81 \times 10^{-6} = 8.281 \times 10^{-5}$
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188
ChemistryEasyMCQMHT CET · 2024
Which of the following equations represents the relation between solubility and solubility product for salt $BA_3$?
A
$S = (K_{sp} / 27)^{1/4}$
B
$S = (27 \times K_{sp})^{1/4}$
C
$S = (K_{sp} / 4)^{1/4}$
D
$S = (4 \times K_{sp})^{1/4}$
Solution
(A) The salt $BA_3$ ionizes in an aqueous solution as follows:
$BA_3(s) \rightleftharpoons B^{3+}(aq) + 3A^{-}(aq)$
Let the solubility of $BA_3$ be $S \ mol/L$.
Then,$[B^{3+}] = S$ and $[A^{-}] = 3S$.
The solubility product constant $(K_{sp})$ is given by:
$K_{sp} = [B^{3+}][A^{-}]^3$
$K_{sp} = (S)(3S)^3$
$K_{sp} = S \times 27S^3 = 27S^4$
Rearranging for $S$:
$S^4 = K_{sp} / 27$
$S = (K_{sp} / 27)^{1/4}$
$BA_3(s) \rightleftharpoons B^{3+}(aq) + 3A^{-}(aq)$
Let the solubility of $BA_3$ be $S \ mol/L$.
Then,$[B^{3+}] = S$ and $[A^{-}] = 3S$.
The solubility product constant $(K_{sp})$ is given by:
$K_{sp} = [B^{3+}][A^{-}]^3$
$K_{sp} = (S)(3S)^3$
$K_{sp} = S \times 27S^3 = 27S^4$
Rearranging for $S$:
$S^4 = K_{sp} / 27$
$S = (K_{sp} / 27)^{1/4}$
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189
ChemistryMediumMCQMHT CET · 2024
Calculate the solubility product of sparingly soluble salt $BA$ at $25^{\circ} C$ if its solubility is $7.2 \times 10^{-7} \ mol \ dm^{-3}$ at same temperature.
A
$4.810 \times 10^{-13}$
B
$5.184 \times 10^{-13}$
C
$6.454 \times 10^{-13}$
D
$5.925 \times 10^{-13}$
Solution
(B) The dissociation of the sparingly soluble salt $BA$ is given by:
$BA_{(s)} \rightleftharpoons B_{(aq)}^{+} + A_{(aq)}^{-}$
For a salt of the type $BA$,the solubility product $K_{sp}$ is related to solubility $S$ by the formula:
$K_{sp} = [B^+][A^-] = S \times S = S^2$
Given,solubility $S = 7.2 \times 10^{-7} \ mol \ dm^{-3}$.
Substituting the value of $S$ in the formula:
$K_{sp} = (7.2 \times 10^{-7})^2$
$K_{sp} = 51.84 \times 10^{-14} = 5.184 \times 10^{-13}$
$BA_{(s)} \rightleftharpoons B_{(aq)}^{+} + A_{(aq)}^{-}$
For a salt of the type $BA$,the solubility product $K_{sp}$ is related to solubility $S$ by the formula:
$K_{sp} = [B^+][A^-] = S \times S = S^2$
Given,solubility $S = 7.2 \times 10^{-7} \ mol \ dm^{-3}$.
Substituting the value of $S$ in the formula:
$K_{sp} = (7.2 \times 10^{-7})^2$
$K_{sp} = 51.84 \times 10^{-14} = 5.184 \times 10^{-13}$
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190
ChemistryMediumMCQMHT CET · 2024
Calculate the solubility of sparingly soluble salt $BA$ in $mol \ dm^{-3}$ at $300 \ K$ if its solubility product is $4.9 \times 10^{-9}$ at same temperature.
A
$5.72 \times 10^{-5}$
B
$6.40 \times 10^{-5}$
C
$7.00 \times 10^{-5}$
D
$7.81 \times 10^{-5}$
Solution
(C) The dissociation of the salt $BA$ is given by: $BA_{(s)} \rightleftharpoons B_{(aq)}^{+} + A_{(aq)}^{-}$
Let the solubility of the salt be $S \ mol \ dm^{-3}$.
Then,$[B^{+}] = S$ and $[A^{-}] = S$.
The solubility product constant $K_{sp}$ is given by: $K_{sp} = [B^{+}][A^{-}] = S \times S = S^2$.
Given $K_{sp} = 4.9 \times 10^{-9}$.
Therefore,$S^2 = 4.9 \times 10^{-9} = 49 \times 10^{-10}$.
Taking the square root on both sides: $S = \sqrt{49 \times 10^{-10}} = 7.00 \times 10^{-5} \ mol \ dm^{-3}$.
Let the solubility of the salt be $S \ mol \ dm^{-3}$.
Then,$[B^{+}] = S$ and $[A^{-}] = S$.
The solubility product constant $K_{sp}$ is given by: $K_{sp} = [B^{+}][A^{-}] = S \times S = S^2$.
Given $K_{sp} = 4.9 \times 10^{-9}$.
Therefore,$S^2 = 4.9 \times 10^{-9} = 49 \times 10^{-10}$.
Taking the square root on both sides: $S = \sqrt{49 \times 10^{-10}} = 7.00 \times 10^{-5} \ mol \ dm^{-3}$.
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191
ChemistryDifficultMCQMHT CET · 2024
Calculate solubility $(mol \ dm^{-3})$ of a sparingly soluble electrolyte $AB$ at $298 \ K$ if its solubility product is $1.6 \times 10^{-5}$?
A
$1.6 \times 10^{-3}$
B
$2.5 \times 10^{-3}$
C
$4.0 \times 10^{-3}$
D
$8.0 \times 10^{-3}$
Solution
(C) For a sparingly soluble salt $AB$,the dissociation equilibrium is:
$AB_{(s)} \rightleftharpoons A_{(aq)}^{+} + B_{(aq)}^{-}$
Let the solubility be $S \ mol \ dm^{-3}$.
Then,$[A^{+}] = S$ and $[B^{-}] = S$.
The solubility product constant $K_{sp}$ is given by:
$K_{sp} = [A^{+}][B^{-}] = S \times S = S^2$
Given $K_{sp} = 1.6 \times 10^{-5}$.
Therefore,$S^2 = 1.6 \times 10^{-5} = 16 \times 10^{-6}$.
Taking the square root on both sides:
$S = \sqrt{16 \times 10^{-6}} = 4.0 \times 10^{-3} \ mol \ dm^{-3}$.
$AB_{(s)} \rightleftharpoons A_{(aq)}^{+} + B_{(aq)}^{-}$
Let the solubility be $S \ mol \ dm^{-3}$.
Then,$[A^{+}] = S$ and $[B^{-}] = S$.
The solubility product constant $K_{sp}$ is given by:
$K_{sp} = [A^{+}][B^{-}] = S \times S = S^2$
Given $K_{sp} = 1.6 \times 10^{-5}$.
Therefore,$S^2 = 1.6 \times 10^{-5} = 16 \times 10^{-6}$.
Taking the square root on both sides:
$S = \sqrt{16 \times 10^{-6}} = 4.0 \times 10^{-3} \ mol \ dm^{-3}$.
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192
ChemistryEasyMCQMHT CET · 2024
Which from the following equations represents the relation between solubility ($S$ in $mol \ L^{-1}$) and solubility product $(K_{sp})$ for a salt $B_3A_2$?
A
$S = \left( \frac{K_{sp}}{108} \right)^{\frac{1}{5}}$
B
$S = \left( 108 \times K_{sp} \right)^{\frac{1}{5}}$
C
$S = \left( \frac{K_{sp}}{27} \right)^{\frac{1}{5}}$
D
$S = \left( 27 \times K_{sp} \right)^{\frac{1}{5}}$
Solution
(A) The dissociation of the salt $B_3A_2$ is given by:
$B_3A_{2(s)} \rightleftharpoons 3B^{2+} + 2A^{3-}$
Let the solubility be $S \ mol \ L^{-1}$.
Then,$[B^{2+}] = 3S$ and $[A^{3-}] = 2S$.
The solubility product expression is:
$K_{sp} = [B^{2+}]^3 [A^{3-}]^2$
Substituting the values:
$K_{sp} = (3S)^3 (2S)^2$
$K_{sp} = (27S^3) \times (4S^2)$
$K_{sp} = 108S^5$
Solving for $S$:
$S^5 = \frac{K_{sp}}{108}$
$S = \left( \frac{K_{sp}}{108} \right)^{\frac{1}{5}}$
$B_3A_{2(s)} \rightleftharpoons 3B^{2+} + 2A^{3-}$
Let the solubility be $S \ mol \ L^{-1}$.
Then,$[B^{2+}] = 3S$ and $[A^{3-}] = 2S$.
The solubility product expression is:
$K_{sp} = [B^{2+}]^3 [A^{3-}]^2$
Substituting the values:
$K_{sp} = (3S)^3 (2S)^2$
$K_{sp} = (27S^3) \times (4S^2)$
$K_{sp} = 108S^5$
Solving for $S$:
$S^5 = \frac{K_{sp}}{108}$
$S = \left( \frac{K_{sp}}{108} \right)^{\frac{1}{5}}$
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193
ChemistryMediumMCQMHT CET · 2024
Calculate the solubility in $mol \ dm^{-3}$ of sparingly soluble salt $BA$ if its solubility product is $4.9 \times 10^{-13}$ at the same temperature.
A
$7.0 \times 10^{-7}$
B
$7.5 \times 10^{-7}$
C
$8.0 \times 10^{-7}$
D
$4.9 \times 10^{-7}$
Solution
(A) For a sparingly soluble salt $BA$,the dissociation equilibrium is represented as:
$BA_{(s)} \rightleftharpoons B_{(aq)}^{+} + A_{(aq)}^{-}$
Let the solubility of $BA$ be $S \ mol \ dm^{-3}$.
Then,$[B^{+}] = S$ and $[A^{-}] = S$.
The solubility product constant $K_{sp}$ is given by:
$K_{sp} = [B^{+}][A^{-}] = S \times S = S^2$
Given $K_{sp} = 4.9 \times 10^{-13}$.
Therefore,$S^2 = 4.9 \times 10^{-13} = 49 \times 10^{-14}$.
Taking the square root on both sides:
$S = \sqrt{49 \times 10^{-14}} = 7.0 \times 10^{-7} \ mol \ dm^{-3}$.
$BA_{(s)} \rightleftharpoons B_{(aq)}^{+} + A_{(aq)}^{-}$
Let the solubility of $BA$ be $S \ mol \ dm^{-3}$.
Then,$[B^{+}] = S$ and $[A^{-}] = S$.
The solubility product constant $K_{sp}$ is given by:
$K_{sp} = [B^{+}][A^{-}] = S \times S = S^2$
Given $K_{sp} = 4.9 \times 10^{-13}$.
Therefore,$S^2 = 4.9 \times 10^{-13} = 49 \times 10^{-14}$.
Taking the square root on both sides:
$S = \sqrt{49 \times 10^{-14}} = 7.0 \times 10^{-7} \ mol \ dm^{-3}$.
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194
ChemistryMediumMCQMHT CET · 2024
Calculate the solubility product of sparingly soluble salt $BA$ at $27^{\circ} C$ if its solubility is $1.8 \times 10^{-5} \ mol \ dm^{-3}$ at same temperature.
A
$3.24 \times 10^{-10}$
B
$2.44 \times 10^{-10}$
C
$1.64 \times 10^{-10}$
D
$4.00 \times 10^{-10}$
Solution
(A) The dissociation of the sparingly soluble salt $BA$ is given by:
$BA_{(s)} \rightleftharpoons B_{(aq)}^{+} + A_{(aq)}^{-}$
For a salt of the type $BA$,the solubility product $K_{sp}$ is related to solubility $S$ by the expression:
$K_{sp} = [B^{+}][A^{-}] = S \times S = S^2$
Given the solubility $S = 1.8 \times 10^{-5} \ mol \ dm^{-3}$,we calculate $K_{sp}$ as:
$K_{sp} = (1.8 \times 10^{-5})^2 = 3.24 \times 10^{-10} \ mol^2 \ dm^{-6}$
$BA_{(s)} \rightleftharpoons B_{(aq)}^{+} + A_{(aq)}^{-}$
For a salt of the type $BA$,the solubility product $K_{sp}$ is related to solubility $S$ by the expression:
$K_{sp} = [B^{+}][A^{-}] = S \times S = S^2$
Given the solubility $S = 1.8 \times 10^{-5} \ mol \ dm^{-3}$,we calculate $K_{sp}$ as:
$K_{sp} = (1.8 \times 10^{-5})^2 = 3.24 \times 10^{-10} \ mol^2 \ dm^{-6}$
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195
ChemistryMediumMCQMHT CET · 2024
What is the bond angle $\angle F-B-F$ in $BF_3$ (in $^{\circ}$)?
A
$107$
B
$104.5$
C
$120$
D
$109.5$
Solution
(C) The $BF_3$ molecule has a trigonal planar geometry.
In this structure,the central Boron atom is $sp^2$ hybridized.
Due to the trigonal planar arrangement,the bond angle between any two $F-B-F$ bonds is $120^{\circ}$.
In this structure,the central Boron atom is $sp^2$ hybridized.
Due to the trigonal planar arrangement,the bond angle between any two $F-B-F$ bonds is $120^{\circ}$.
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196
ChemistryMediumMCQMHT CET · 2024
Which of the following compounds is the most covalent?
A
$SbCl_3$
B
$PbCl_2$
C
$SnCl_4$
D
$SnCl_2$
Solution
(C) According to Fajan's rule,the covalent character of a compound increases with the increase in the oxidation state of the metal cation.
Higher oxidation state leads to higher polarising power of the cation,which results in greater covalent character.
Comparing the oxidation states of the metals in the given compounds:
$SbCl_3$: $Sb$ is in $+3$ oxidation state.
$PbCl_2$: $Pb$ is in $+2$ oxidation state.
$SnCl_4$: $Sn$ is in $+4$ oxidation state.
$SnCl_2$: $Sn$ is in $+2$ oxidation state.
Since $Sn$ in $SnCl_4$ has the highest oxidation state $(+4)$,it has the highest polarising power and thus $SnCl_4$ is the most covalent compound among the given options.
Higher oxidation state leads to higher polarising power of the cation,which results in greater covalent character.
Comparing the oxidation states of the metals in the given compounds:
$SbCl_3$: $Sb$ is in $+3$ oxidation state.
$PbCl_2$: $Pb$ is in $+2$ oxidation state.
$SnCl_4$: $Sn$ is in $+4$ oxidation state.
$SnCl_2$: $Sn$ is in $+2$ oxidation state.
Since $Sn$ in $SnCl_4$ has the highest oxidation state $(+4)$,it has the highest polarising power and thus $SnCl_4$ is the most covalent compound among the given options.
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197
ChemistryEasyMCQMHT CET · 2024
What is the number of moles of $Cl$ atoms and $N$ atoms respectively present in $n$ moles of tear gas?
A
$3n$ $Cl$ and $n$ $N$
B
$2n$ $Cl$ and $2n$ $N$
C
$n$ $Cl$ and $n$ $N$
D
$n$ $Cl$ and $2n$ $N$
Solution
(A) The molecular formula of tear gas $(CCl_3NO_2)$ contains $3$ atoms of $Cl$ and $1$ atom of $N$ per molecule.
Therefore,in $n$ moles of tear gas,the number of moles of $Cl$ atoms is $3n$ and the number of moles of $N$ atoms is $n$.
Therefore,in $n$ moles of tear gas,the number of moles of $Cl$ atoms is $3n$ and the number of moles of $N$ atoms is $n$.
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198
ChemistryDifficultMCQMHT CET · 2024
What is the value of electronegativity of oxygen?
A
$4$
B
$3.5$
C
$2.48$
D
$3.2$
Solution
(B) According to the Pauling scale,the electronegativity of oxygen is $3.5$.
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199
ChemistryDifficultMCQMHT CET · 2024
What is the $O-O$ bond length in the ozone $(O_3)$ molecule (in $pm$)?
A
$128$
B
$117$
C
$107$
D
$134$
Solution
(A) The ozone $(O_3)$ molecule is a resonance hybrid of two canonical structures.
Due to resonance, both $O-O$ bonds in the ozone molecule are equivalent.
The experimental bond length for both $O-O$ bonds in the resonance hybrid of ozone is $128 \ pm$.
Due to resonance, both $O-O$ bonds in the ozone molecule are equivalent.
The experimental bond length for both $O-O$ bonds in the resonance hybrid of ozone is $128 \ pm$.
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200
ChemistryMediumMCQMHT CET · 2024
Identify the bond angle $O-S-O$ in the $SO_2$ molecule. (in $^{\circ}$)
A
$119.5$
B
$180$
C
$109$
D
$107.5$
Solution
(A) The $SO_2$ molecule has a bent or $V$-shaped geometry due to the presence of one lone pair on the sulfur atom.
According to $VSEPR$ theory,the lone pair-bond pair repulsion is greater than the bond pair-bond pair repulsion,which causes the bond angle to decrease from the ideal $120^{\circ}$ (for $sp^2$ hybridization) to approximately $119.3^{\circ}$.
Among the given options,$119.5^{\circ}$ is the closest value to the experimental bond angle.
According to $VSEPR$ theory,the lone pair-bond pair repulsion is greater than the bond pair-bond pair repulsion,which causes the bond angle to decrease from the ideal $120^{\circ}$ (for $sp^2$ hybridization) to approximately $119.3^{\circ}$.
Among the given options,$119.5^{\circ}$ is the closest value to the experimental bond angle.
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201
ChemistryEasyMCQMHT CET · 2024
Half life of a zero order reaction is directly proportional to $\qquad$
A
temperature
B
rate constant
C
amount of product formed
D
initial concentration of reactant
Solution
(D) The half-life $(t_{1/2})$ of a zero-order reaction is given by the formula: $t_{1/2} = \frac{[A]_0}{2k}$
where $[A]_0$ is the initial concentration of the reactant and $k$ is the rate constant.
From the formula,it is clear that $t_{1/2} \propto [A]_0$.
Therefore,the half-life of a zero-order reaction is directly proportional to the initial concentration of the reactant.
where $[A]_0$ is the initial concentration of the reactant and $k$ is the rate constant.
From the formula,it is clear that $t_{1/2} \propto [A]_0$.
Therefore,the half-life of a zero-order reaction is directly proportional to the initial concentration of the reactant.
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202
ChemistryEasyMCQMHT CET · 2024
What is the rate constant of a first-order reaction if $60 \%$ of the reactant decomposes in $45 \ minutes$?
A
$0.010 \ minute^{-1}$
B
$0.015 \ minute^{-1}$
C
$0.020 \ minute^{-1}$
D
$0.025 \ minute^{-1}$
Solution
(C) For a first-order reaction,the rate constant $k$ is given by the formula: $k = \frac{2.303}{t} \log_{10} \frac{[A]_0}{[A]_t}$.
Given that $60 \%$ of the reactant decomposes,the remaining concentration $[A]_t$ is $100 - 60 = 40$ if the initial concentration $[A]_0 = 100$.
Substituting the values: $k = \frac{2.303}{45} \log_{10} \frac{100}{40}$.
$k = \frac{2.303}{45} \log_{10} (2.5)$.
Since $\log_{10} (2.5) \approx 0.3979$,we get $k = \frac{2.303 \times 0.3979}{45}$.
$k \approx 0.020 \ minute^{-1}$.
Given that $60 \%$ of the reactant decomposes,the remaining concentration $[A]_t$ is $100 - 60 = 40$ if the initial concentration $[A]_0 = 100$.
Substituting the values: $k = \frac{2.303}{45} \log_{10} \frac{100}{40}$.
$k = \frac{2.303}{45} \log_{10} (2.5)$.
Since $\log_{10} (2.5) \approx 0.3979$,we get $k = \frac{2.303 \times 0.3979}{45}$.
$k \approx 0.020 \ minute^{-1}$.
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203
ChemistryEasyMCQMHT CET · 2024
For a zero order reaction $A \longrightarrow \text{product}$,the concentration of $A$ decreases from $0.8 \ mol \ dm^{-3}$ to $0.2 \ mol \ dm^{-3}$ in $6 \ minute$. What is the rate constant of the reaction?
A
$0.01 \ mol \ dm^{-3} \ minute^{-1}$
B
$1.0 \ mol \ dm^{-3} \ minute^{-1}$
C
$0.1 \ mol \ dm^{-3} \ minute^{-1}$
D
$1.66 \ mol \ dm^{-3} \ minute^{-1}$
Solution
(C) For a zero order reaction,the rate constant $k$ is given by the formula:
$k = \frac{[A]_0 - [A]_t}{t}$
Given:
$[A]_0 = 0.8 \ mol \ dm^{-3}$
$[A]_t = 0.2 \ mol \ dm^{-3}$
$t = 6 \ minute$
Substituting the values:
$k = \frac{0.8 - 0.2}{6} \ mol \ dm^{-3} \ minute^{-1}$
$k = \frac{0.6}{6} \ mol \ dm^{-3} \ minute^{-1}$
$k = 0.1 \ mol \ dm^{-3} \ minute^{-1}$
$k = \frac{[A]_0 - [A]_t}{t}$
Given:
$[A]_0 = 0.8 \ mol \ dm^{-3}$
$[A]_t = 0.2 \ mol \ dm^{-3}$
$t = 6 \ minute$
Substituting the values:
$k = \frac{0.8 - 0.2}{6} \ mol \ dm^{-3} \ minute^{-1}$
$k = \frac{0.6}{6} \ mol \ dm^{-3} \ minute^{-1}$
$k = 0.1 \ mol \ dm^{-3} \ minute^{-1}$
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204
ChemistryEasyMCQMHT CET · 2024
The half-life of a first-order reaction is $1 \text{ hour}$. What fraction of the reactant will remain after $3 \text{ hours}$?
A
$\frac{1}{8}$
B
$\frac{1}{9}$
C
$\frac{1}{16}$
D
$\frac{1}{64}$
Solution
(A) For a first-order reaction,the amount remaining after $n$ half-lives is given by the formula: $\frac{[A]_t}{[A]_0} = (\frac{1}{2})^n$.
Here,the total time $t = 3 \text{ hours}$ and the half-life $t_{1/2} = 1 \text{ hour}$.
The number of half-lives $n = \frac{t}{t_{1/2}} = \frac{3}{1} = 3$.
Therefore,the fraction remaining is $(\frac{1}{2})^3 = \frac{1}{8}$.
Here,the total time $t = 3 \text{ hours}$ and the half-life $t_{1/2} = 1 \text{ hour}$.
The number of half-lives $n = \frac{t}{t_{1/2}} = \frac{3}{1} = 3$.
Therefore,the fraction remaining is $(\frac{1}{2})^3 = \frac{1}{8}$.
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205
ChemistryDifficultMCQMHT CET · 2024
Rate of a first order reaction is $1.5 \times 10^{-2} \ mol \ L^{-1} \ minute^{-1}$ at $0.5 \ M$ concentration of reactant,calculate half life of reaction.
A
$0.383 \ minute$
B
$7.53 \ minute$
C
$8.73 \ minute$
D
$23.1 \ minute$
Solution
(D) For a $1^{st}$ order reaction,the rate law is given by: $\text{Rate} = k[A]$.
Given: $\text{Rate} = 1.5 \times 10^{-2} \ mol \ L^{-1} \ minute^{-1}$ and $[A] = 0.5 \ M$.
Substituting the values: $1.5 \times 10^{-2} = k(0.5)$.
Calculating the rate constant: $k = \frac{1.5 \times 10^{-2}}{0.5} = 0.03 \ minute^{-1}$.
The half-life $(t_{1/2})$ for a $1^{st}$ order reaction is calculated as: $t_{1/2} = \frac{0.693}{k}$.
$t_{1/2} = \frac{0.693}{0.03 \ minute^{-1}} = 23.1 \ minute$.
Given: $\text{Rate} = 1.5 \times 10^{-2} \ mol \ L^{-1} \ minute^{-1}$ and $[A] = 0.5 \ M$.
Substituting the values: $1.5 \times 10^{-2} = k(0.5)$.
Calculating the rate constant: $k = \frac{1.5 \times 10^{-2}}{0.5} = 0.03 \ minute^{-1}$.
The half-life $(t_{1/2})$ for a $1^{st}$ order reaction is calculated as: $t_{1/2} = \frac{0.693}{k}$.
$t_{1/2} = \frac{0.693}{0.03 \ minute^{-1}} = 23.1 \ minute$.
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206
ChemistryEasyMCQMHT CET · 2024
Initial concentration of reactant in a first order reaction is $0.08 \text{ mol dm}^{-3}$. What concentration would remain after $40 \text{ minutes}$? (Given $\frac{[A]_0}{[A]_t} = 5.00$)
A
$0.008 \text{ mol dm}^{-3}$
B
$0.08 \text{ mol dm}^{-3}$
C
$0.016 \text{ mol dm}^{-3}$
D
$0.032 \text{ mol dm}^{-3}$
Solution
(C) Given that the initial concentration $[A]_0 = 0.08 \text{ mol dm}^{-3}$.
We are given the ratio $\frac{[A]_0}{[A]_t} = 5.00$.
To find the remaining concentration $[A]_t$,we rearrange the equation:
$[A]_t = \frac{[A]_0}{5.00} = \frac{0.08 \text{ mol dm}^{-3}}{5.00} = 0.016 \text{ mol dm}^{-3}$.
Thus,the concentration remaining after $40 \text{ minutes}$ is $0.016 \text{ mol dm}^{-3}$.
We are given the ratio $\frac{[A]_0}{[A]_t} = 5.00$.
To find the remaining concentration $[A]_t$,we rearrange the equation:
$[A]_t = \frac{[A]_0}{5.00} = \frac{0.08 \text{ mol dm}^{-3}}{5.00} = 0.016 \text{ mol dm}^{-3}$.
Thus,the concentration remaining after $40 \text{ minutes}$ is $0.016 \text{ mol dm}^{-3}$.
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207
ChemistryDifficultMCQMHT CET · 2024
The rate constant of the reaction $2 NO_2Cl_{(g)} \longrightarrow 2 NO_{2(g)} + Cl_{2(g)}$ is $4.7672 \text{ minute}^{-1}$. Calculate the half-life of the reaction.
A
$0.0727 \text{ minute}$
B
$0.1454 \text{ minute}$
C
$0.2181 \text{ minute}$
D
$0.4362 \text{ minute}$
Solution
(B) The given reaction is a first-order reaction because the unit of the rate constant is $\text{minute}^{-1}$.
For a first-order reaction,the half-life $(t_{1/2})$ is given by the formula:
$t_{1/2} = \frac{0.693}{k}$
Given $k = 4.7672 \text{ minute}^{-1}$.
$t_{1/2} = \frac{0.693}{4.7672 \text{ minute}^{-1}} = 0.1454 \text{ minute}$.
For a first-order reaction,the half-life $(t_{1/2})$ is given by the formula:
$t_{1/2} = \frac{0.693}{k}$
Given $k = 4.7672 \text{ minute}^{-1}$.
$t_{1/2} = \frac{0.693}{4.7672 \text{ minute}^{-1}} = 0.1454 \text{ minute}$.
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208
ChemistryEasyMCQMHT CET · 2024
$A$ zero order reaction has a half-life time of $0.2 \ min$. If the initial concentration of the reactant is $0.2 \ mol \ dm^{-3}$,find the rate constant.
A
$0.2 \ mol \ dm^{-3} \ min^{-1}$
B
$0.5 \ mol \ dm^{-3} \ min^{-1}$
C
$1.4 \ mol \ dm^{-3} \ min^{-1}$
D
$6.0 \ mol \ dm^{-3} \ min^{-1}$
Solution
(B) For a zero order reaction,the half-life is given by the formula: $t_{1/2} = \frac{[A]_0}{2k}$
Given: $t_{1/2} = 0.2 \ min$ and $[A]_0 = 0.2 \ mol \ dm^{-3}$.
Substituting the values: $0.2 \ min = \frac{0.2 \ mol \ dm^{-3}}{2k}$
$2k = \frac{0.2 \ mol \ dm^{-3}}{0.2 \ min} = 1 \ mol \ dm^{-3} \ min^{-1}$
$k = 0.5 \ mol \ dm^{-3} \ min^{-1}$
Given: $t_{1/2} = 0.2 \ min$ and $[A]_0 = 0.2 \ mol \ dm^{-3}$.
Substituting the values: $0.2 \ min = \frac{0.2 \ mol \ dm^{-3}}{2k}$
$2k = \frac{0.2 \ mol \ dm^{-3}}{0.2 \ min} = 1 \ mol \ dm^{-3} \ min^{-1}$
$k = 0.5 \ mol \ dm^{-3} \ min^{-1}$
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209
ChemistryEasyMCQMHT CET · 2024
In a first order reaction,$60 \%$ of the reactant converts into product in $45 \ minutes$. Calculate the rate constant of the reaction.
A
$0.0102 \ minute^{-1}$
B
$0.0204 \ minute^{-1}$
C
$0.0306 \ minute^{-1}$
D
$0.0408 \ minute^{-1}$
Solution
(B) For a first order reaction,the rate constant $k$ is given by the formula: $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$
Given that $60 \%$ of the reactant is converted,the remaining concentration $[A]_t$ is $100 - 60 = 40 \%$ of the initial concentration $[A]_0$.
So,$[A]_0 = 100$ and $[A]_t = 40$.
Substituting the values: $k = \frac{2.303}{45} \log \frac{100}{40}$
$k = \frac{2.303}{45} \log(2.5)$
$k = \frac{2.303}{45} \times 0.3979$
$k \approx 0.02036 \ minute^{-1} \approx 0.0204 \ minute^{-1}$.
Given that $60 \%$ of the reactant is converted,the remaining concentration $[A]_t$ is $100 - 60 = 40 \%$ of the initial concentration $[A]_0$.
So,$[A]_0 = 100$ and $[A]_t = 40$.
Substituting the values: $k = \frac{2.303}{45} \log \frac{100}{40}$
$k = \frac{2.303}{45} \log(2.5)$
$k = \frac{2.303}{45} \times 0.3979$
$k \approx 0.02036 \ minute^{-1} \approx 0.0204 \ minute^{-1}$.
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210
ChemistryEasyMCQMHT CET · 2024
What is the order of the following reaction: $2 H_2 O_{2(g)} \longrightarrow 2 H_2 O_{(l)} + O_{2(g)}$?
A
$1$
B
$0$
C
$3$
D
$2$
Solution
(D) The decomposition of hydrogen peroxide $(H_2 O_2)$ is a well-studied chemical reaction.
Experimentally,the rate of decomposition of $H_2 O_2$ is found to be proportional to the square of the concentration of $H_2 O_2$.
The rate law is given by: $\text{Rate} = k [H_2 O_2]^2$.
Since the exponent of the concentration term in the rate law is $2$,the order of the reaction is $2$.
Experimentally,the rate of decomposition of $H_2 O_2$ is found to be proportional to the square of the concentration of $H_2 O_2$.
The rate law is given by: $\text{Rate} = k [H_2 O_2]^2$.
Since the exponent of the concentration term in the rate law is $2$,the order of the reaction is $2$.
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211
ChemistryEasyMCQMHT CET · 2024
What is the value of the slope if $\log_{10} K$ ($y$-axis) is plotted versus $1/T$ ($x$-axis) for the Arrhenius equation?
A
$\log_{10} A$
B
$\frac{2.303 R}{E_a}$
C
$\frac{-E_a}{2.303 R}$
D
$-\log_{10} A$
Solution
(C) The Arrhenius equation is given by $k = A e^{-E_a / RT}$.
Taking the natural logarithm on both sides:
$\ln k = -\frac{E_a}{RT} + \ln A$
To convert this into base $10$ logarithm,we divide by $2.303$:
$\log_{10} k = -\frac{E_a}{2.303 R} \cdot \frac{1}{T} + \log_{10} A$
Comparing this with the equation of a straight line $y = mx + c$,where $y = \log_{10} k$ and $x = 1/T$,the slope $m$ is equal to $-\frac{E_a}{2.303 R}$.
Taking the natural logarithm on both sides:
$\ln k = -\frac{E_a}{RT} + \ln A$
To convert this into base $10$ logarithm,we divide by $2.303$:
$\log_{10} k = -\frac{E_a}{2.303 R} \cdot \frac{1}{T} + \log_{10} A$
Comparing this with the equation of a straight line $y = mx + c$,where $y = \log_{10} k$ and $x = 1/T$,the slope $m$ is equal to $-\frac{E_a}{2.303 R}$.
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212
ChemistryMediumMCQMHT CET · 2024
In the Arrhenius plot of $\log_{10} k$ versus $1 / T$,find the value of the intercept on the $y$-axis.
A
$\log_{10} A$
B
$\frac{-E_a}{R}$
C
$\ln k$
D
$\frac{R}{E_a}$
Solution
(A) The Arrhenius equation is given by: $k = A e^{-E_a / RT}$.
Taking the natural logarithm on both sides: $\ln k = -\frac{E_a}{RT} + \ln A$.
Converting to base $10$ logarithm: $\log_{10} k = -\frac{E_a}{2.303 R} \left(\frac{1}{T}\right) + \log_{10} A$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = \log_{10} k$,$x = \frac{1}{T}$,$m = -\frac{E_a}{2.303 R}$,and the intercept $c = \log_{10} A$.
Therefore,the intercept on the $y$-axis is $\log_{10} A$.
Taking the natural logarithm on both sides: $\ln k = -\frac{E_a}{RT} + \ln A$.
Converting to base $10$ logarithm: $\log_{10} k = -\frac{E_a}{2.303 R} \left(\frac{1}{T}\right) + \log_{10} A$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = \log_{10} k$,$x = \frac{1}{T}$,$m = -\frac{E_a}{2.303 R}$,and the intercept $c = \log_{10} A$.
Therefore,the intercept on the $y$-axis is $\log_{10} A$.
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213
ChemistryMediumMCQMHT CET · 2024
Which of the following is true for a reaction as per collision theory?
A
Every collision between reactants leads to chemical reaction.
B
It may be expected that rate of reaction is equal to rate of collision.
C
For gas phase reactions,the number of collisions is far less compared to observed rate.
D
The colliding molecules do not need proper orientation.
Solution
(B) According to collision theory,not every collision between reactant molecules leads to a chemical reaction. Only those collisions that possess energy greater than the threshold energy and have proper orientation result in a product.
For gas phase reactions,the calculated number of collisions is much higher than the observed rate of reaction.
Therefore,the statement that it may be expected that the rate of reaction is equal to the rate of collision is the correct premise for the theory's development before accounting for the steric factor and activation energy.
For gas phase reactions,the calculated number of collisions is much higher than the observed rate of reaction.
Therefore,the statement that it may be expected that the rate of reaction is equal to the rate of collision is the correct premise for the theory's development before accounting for the steric factor and activation energy.
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214
ChemistryMediumMCQMHT CET · 2024
Identify an aldehyde used in margarine and in food for its buttery odour.
A
Benzaldehyde
B
Butyraldehyde
C
Cinnamaldehyde
D
Oxaldehyde
Solution
(B) The correct answer is $B$ (Butyraldehyde).
Butyraldehyde (also known as butanal) is an aldehyde that has a characteristic buttery odor.
It is commonly used in the food industry to impart a buttery flavor,and it is also found in margarine to enhance its aroma.
Here is a quick look at the other options:
- $A$ Benzaldehyde: This aldehyde has an almond-like aroma and is used in flavorings and perfumes,but not for a buttery odor.
- $C$ Cinnamaldehyde: This aldehyde is responsible for the characteristic flavor and aroma of cinnamon,not butter.
- $D$ Oxaldehyde: This is not a common aldehyde and does not fit the description related to buttery odors.
Thus,Butyraldehyde is the aldehyde associated with the buttery scent in margarine and other food products.
Butyraldehyde (also known as butanal) is an aldehyde that has a characteristic buttery odor.
It is commonly used in the food industry to impart a buttery flavor,and it is also found in margarine to enhance its aroma.
Here is a quick look at the other options:
- $A$ Benzaldehyde: This aldehyde has an almond-like aroma and is used in flavorings and perfumes,but not for a buttery odor.
- $C$ Cinnamaldehyde: This aldehyde is responsible for the characteristic flavor and aroma of cinnamon,not butter.
- $D$ Oxaldehyde: This is not a common aldehyde and does not fit the description related to buttery odors.
Thus,Butyraldehyde is the aldehyde associated with the buttery scent in margarine and other food products.
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215
ChemistryEasyMCQMHT CET · 2024
Which of the following is the largest size nanomaterial?
A
Water (molecular level)
B
Glucose (molecular level)
C
Virus
D
Bacteria
Solution
(D) The size ranges are as follows:
- Water and Glucose are at the molecular scale,typically ranging from $0.1$ to $1 \ nm$.
- Viruses typically range from $20$ to $300 \ nm$.
- Bacteria are much larger,typically ranging from $0.2$ to $10 \ \mu m$ ($200$ to $10,000 \ nm$).
Therefore,among the given options,bacteria are the largest.
- Water and Glucose are at the molecular scale,typically ranging from $0.1$ to $1 \ nm$.
- Viruses typically range from $20$ to $300 \ nm$.
- Bacteria are much larger,typically ranging from $0.2$ to $10 \ \mu m$ ($200$ to $10,000 \ nm$).
Therefore,among the given options,bacteria are the largest.
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216
ChemistryDifficultMCQMHT CET · 2024
What type of information is collected using $FTIR$ (Fourier transform infrared) spectroscopy?
A
Morphology of nanomaterial
B
Absorption of functional group
C
Geometry of particles
D
Particle size
Solution
(B) $1$. Morphology of nanomaterial: $FTIR$ does not provide information about the morphology (shape,structure) of nanomaterials. Techniques like $SEM$ or $TEM$ are used for this.
$2$. Absorption of functional group: $FTIR$ identifies functional groups by detecting characteristic absorption bands in the infrared region,such as those for $-OH$,$-CH$,$-NH$,and $-CO$ groups. This is the correct information collected by $FTIR$.
$3$. Geometry of particles: $FTIR$ does not give information about particle geometry (e.g.,spherical or cubic shapes).
$4$. Particle size: $FTIR$ cannot determine particle size. Techniques like $DLS$ (Dynamic Light Scattering) or electron microscopy are used for this.
$2$. Absorption of functional group: $FTIR$ identifies functional groups by detecting characteristic absorption bands in the infrared region,such as those for $-OH$,$-CH$,$-NH$,and $-CO$ groups. This is the correct information collected by $FTIR$.
$3$. Geometry of particles: $FTIR$ does not give information about particle geometry (e.g.,spherical or cubic shapes).
$4$. Particle size: $FTIR$ cannot determine particle size. Techniques like $DLS$ (Dynamic Light Scattering) or electron microscopy are used for this.
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217
ChemistryDifficultMCQMHT CET · 2024
Which from the following compounds is used to prepare adipic acid using enzymes in green technology developed by $Drath$ and $Frost$?
A
Ribose
B
Glucose
C
Ribulose
D
Benzene
Solution
(B) In the green technology developed by $Drath$ and $Frost$,adipic acid is enzymatically synthesized from $Glucose$ rather than $Benzene$,which is carcinogenic.
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218
ChemistryMediumMCQMHT CET · 2024
Identify the element having outer electronic configuration $ns^2 np^5$.
A
$I$
B
$Te$
C
$Ar$
D
$Ne$
Solution
(A) The general outer electronic configuration $ns^2 np^5$ corresponds to the halogen group (Group $17$).
Iodine $(I)$ belongs to the halogen group and has the outer electronic configuration $5s^2 5p^5$,which fits the general form $ns^2 np^5$ where $n=5$.
Therefore,the correct element is $I$.
Iodine $(I)$ belongs to the halogen group and has the outer electronic configuration $5s^2 5p^5$,which fits the general form $ns^2 np^5$ where $n=5$.
Therefore,the correct element is $I$.
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219
ChemistryMediumMCQMHT CET · 2024
Which of the following complexes contains only neutral ligands?
A
$[Co(NH_3)_5(H_2O)]Cl_3$
B
$[Rh(NH_3)_3(SCN)_3]$
C
$[Pt(en)_2(SCN)_2]Cl_2$
D
$[Co(NH_3)_3(NO_2)_3]$
Solution
(A) To determine which complex contains only neutral ligands,we analyze the ligands in each option:
$A$: $[Co(NH_3)_5(H_2O)]Cl_3$ contains $NH_3$ (neutral) and $H_2O$ (neutral).
$B$: $[Rh(NH_3)_3(SCN)_3]$ contains $NH_3$ (neutral) and $SCN^-$ (anionic).
$C$: $[Pt(en)_2(SCN)_2]Cl_2$ contains $en$ (neutral) and $SCN^-$ (anionic).
$D$: $[Co(NH_3)_3(NO_2)_3]$ contains $NH_3$ (neutral) and $NO_2^-$ (anionic).
Therefore,the complex in option $A$ contains only neutral ligands.
$A$: $[Co(NH_3)_5(H_2O)]Cl_3$ contains $NH_3$ (neutral) and $H_2O$ (neutral).
$B$: $[Rh(NH_3)_3(SCN)_3]$ contains $NH_3$ (neutral) and $SCN^-$ (anionic).
$C$: $[Pt(en)_2(SCN)_2]Cl_2$ contains $en$ (neutral) and $SCN^-$ (anionic).
$D$: $[Co(NH_3)_3(NO_2)_3]$ contains $NH_3$ (neutral) and $NO_2^-$ (anionic).
Therefore,the complex in option $A$ contains only neutral ligands.
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220
ChemistryEasyMCQMHT CET · 2024
$EDTA$ (Ethylenediaminetetraacetate ion) is a hexadentate ligand. Identify the number of donor atoms in an $EDTA$ molecule that form coordinate bonds with a central metal atom or ion in a complex.
A
$3$
B
$1$
C
$6$
D
$4$
Solution
(C) $EDTA^{4-}$ (Ethylenediaminetetraacetate ion) is a hexadentate ligand.
It contains $6$ donor atoms: $2$ nitrogen atoms (from the amine groups) and $4$ oxygen atoms (from the four carboxylate groups).
These $6$ donor atoms simultaneously form coordinate bonds with the central metal atom or ion,making it a hexadentate ligand.
Therefore,the number of donor atoms is $6$.
It contains $6$ donor atoms: $2$ nitrogen atoms (from the amine groups) and $4$ oxygen atoms (from the four carboxylate groups).
These $6$ donor atoms simultaneously form coordinate bonds with the central metal atom or ion,making it a hexadentate ligand.
Therefore,the number of donor atoms is $6$.
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221
ChemistryMediumMCQMHT CET · 2024
Find the total number of moles of donor atoms present in one mole of trioxalatocobaltate$(III)$ ion.
A
$6$
B
$3$
C
$12$
D
$4$
Solution
(A) In $[Co(C_2O_4)_3]^{3-}$,there are three oxalate ligands attached to the central cobalt ion.
Since oxalate $(C_2O_4)^{2-}$ is a bidentate ligand,each oxalate ion provides two donor oxygen atoms.
Therefore,$1 \text{ mole of } (C_2O_4)^{2-} \equiv 2 \text{ moles of donor atoms}$.
Thus,$1 \text{ mole of } [Co(C_2O_4)_3]^{3-} \equiv 3 \times 2 = 6 \text{ moles of donor atoms}$.
Since oxalate $(C_2O_4)^{2-}$ is a bidentate ligand,each oxalate ion provides two donor oxygen atoms.
Therefore,$1 \text{ mole of } (C_2O_4)^{2-} \equiv 2 \text{ moles of donor atoms}$.
Thus,$1 \text{ mole of } [Co(C_2O_4)_3]^{3-} \equiv 3 \times 2 = 6 \text{ moles of donor atoms}$.
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222
ChemistryEasyMCQMHT CET · 2024
What is the Effective Atomic Number $(EAN)$ of $Cu$ in $[Cu(NH_3)_4]^{2+}$?
A
$36$
B
$29$
C
$30$
D
$35$
Solution
(D) The formula for $EAN$ is $EAN = Z - X + Y$,where $Z$ is the atomic number of the metal,$X$ is the oxidation state,and $Y$ is the number of electrons donated by the ligands.
For $[Cu(NH_3)_4]^{2+}$:
$Z = 29$ (for $Cu$)
$X = +2$ (as $NH_3$ is neutral,$Cu + 4(0) = +2$)
$Y = 2 \times 4 = 8$ (each $NH_3$ donates $2$ electrons)
$EAN = 29 - 2 + 8 = 35$.
For $[Cu(NH_3)_4]^{2+}$:
$Z = 29$ (for $Cu$)
$X = +2$ (as $NH_3$ is neutral,$Cu + 4(0) = +2$)
$Y = 2 \times 4 = 8$ (each $NH_3$ donates $2$ electrons)
$EAN = 29 - 2 + 8 = 35$.
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223
ChemistryEasyMCQMHT CET · 2024
What is the coordination number of the central metal ion in $[Fe(C_2O_4)_3]^{3-}$?
A
$2$
B
$4$
C
$6$
D
$12$
Solution
(C) The oxalate ligand $(C_2O_4^{2-})$ is a bidentate ligand,meaning it coordinates with the central metal ion through $2$ donor atoms (oxygen atoms) simultaneously.
Since there are $3$ oxalate ligands attached to the central $Fe^{3+}$ ion,the total coordination number is calculated as:
$\text{Coordination Number} = (\text{Number of ligands}) \times (\text{Denticity of ligand})$
$\text{Coordination Number} = 3 \times 2 = 6$.
Since there are $3$ oxalate ligands attached to the central $Fe^{3+}$ ion,the total coordination number is calculated as:
$\text{Coordination Number} = (\text{Number of ligands}) \times (\text{Denticity of ligand})$
$\text{Coordination Number} = 3 \times 2 = 6$.
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224
ChemistryEasyMCQMHT CET · 2024
Which of the following ligands has the highest field strength?
A
$S^{2-}$
B
$OH^{-}$
C
$EDTA^{4-}$
D
$en$
Solution
(D) The spectrochemical series arranges ligands in increasing order of their field strength (crystal field splitting energy):
$I^{-} < Br^{-} < S^{2-} < SCN^{-} < Cl^{-} < F^{-} < OH^{-} < C_2O_4^{2-} < H_2O < NCS^{-} < EDTA^{4-} < NH_3 < en < CN^{-} < CO$.
Comparing the given options:
$S^{2-}$ (weak field),$OH^{-}$ (weak field),$EDTA^{4-}$ (intermediate),and $en$ (strong field).
Among the provided choices,$en$ (ethylenediamine) has the highest field strength.
$I^{-} < Br^{-} < S^{2-} < SCN^{-} < Cl^{-} < F^{-} < OH^{-} < C_2O_4^{2-} < H_2O < NCS^{-} < EDTA^{4-} < NH_3 < en < CN^{-} < CO$.
Comparing the given options:
$S^{2-}$ (weak field),$OH^{-}$ (weak field),$EDTA^{4-}$ (intermediate),and $en$ (strong field).
Among the provided choices,$en$ (ethylenediamine) has the highest field strength.
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225
ChemistryEasyMCQMHT CET · 2024
Which of the following ligands is able to form linkage isomers?
A
$SCN^{-}$
B
$H_2O$
C
$CN^{-}$
D
$C_2O_4^{2-}$
Solution
(A) Ligands that can coordinate through two different donor atoms are called ambidentate ligands.
$SCN^{-}$ is an ambidentate ligand because it can coordinate through either the sulfur atom $(S)$ or the nitrogen atom $(N)$,thus exhibiting linkage isomerism.
$SCN^{-}$ is an ambidentate ligand because it can coordinate through either the sulfur atom $(S)$ or the nitrogen atom $(N)$,thus exhibiting linkage isomerism.
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226
ChemistryEasyMCQMHT CET · 2024
Identify a ligand having two donor atoms but uses a pair of electrons of either donor atom to form a coordinate bond.
A
Aqua
B
Ethylenediamine
C
Sulphato
D
Nitrito
Solution
(D) $NO_2^{-}$ (nitrite) is an ambidentate ligand which has two donor atoms,$N$ and $O$,but uses only one of them at a time to form a coordinate bond.
It is termed as 'nitro' (for $N$-bonded ligand) and 'nitrito' (for $O$-bonded ligand).
It is termed as 'nitro' (for $N$-bonded ligand) and 'nitrito' (for $O$-bonded ligand).
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227
ChemistryEasyMCQMHT CET · 2024
Identify the neutral coordination complex from the following.
A
Pentaamminecobalt$(III)$ sulphate
B
Potassium trioxalatoaluminate$(III)$
C
Diamminedichloroplatinum$(II)$
D
Potassium hexacyanoferrate$(III)$
Solution
(C) neutral coordination complex is one where the coordination sphere carries no net charge.
$1$. Pentaamminecobalt$(III)$ sulphate: $[Co(NH_3)_5(SO_4)]Cl$ or similar ionic forms are cationic complexes.
$2$. Potassium trioxalatoaluminate$(III)$: $K_3[Al(C_2O_4)_3]$ is an anionic complex.
$3$. Diamminedichloroplatinum$(II)$: $[Pt(NH_3)_2Cl_2]$ has a net charge of $0$ $(+2 + 2(0) + 2(-1) = 0)$,making it a neutral complex.
$4$. Potassium hexacyanoferrate$(III)$: $K_3[Fe(CN)_6]$ is an anionic complex.
$1$. Pentaamminecobalt$(III)$ sulphate: $[Co(NH_3)_5(SO_4)]Cl$ or similar ionic forms are cationic complexes.
$2$. Potassium trioxalatoaluminate$(III)$: $K_3[Al(C_2O_4)_3]$ is an anionic complex.
$3$. Diamminedichloroplatinum$(II)$: $[Pt(NH_3)_2Cl_2]$ has a net charge of $0$ $(+2 + 2(0) + 2(-1) = 0)$,making it a neutral complex.
$4$. Potassium hexacyanoferrate$(III)$: $K_3[Fe(CN)_6]$ is an anionic complex.
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228
ChemistryEasyMCQMHT CET · 2024
Identify the heteroleptic complex from the following.
A
$Na_3[Co(NO_2)_6]$
B
$[Fe(H_2O)_5(NCS)]^{2+}$
C
$[Cu(NH_3)_4]^{2+}$
D
$[Co(C_2O_4)_3]^{3-}$
Solution
(B) heteroleptic complex is a coordination compound in which the central metal ion is bonded to more than one type of donor group (ligand).
In the complex $[Fe(H_2O)_5(NCS)]^{2+}$,the central metal ion $Fe^{2+}$ is bonded to two different types of ligands: $H_2O$ and $NCS^-$.
Therefore,it is a heteroleptic complex.
In contrast,the other complexes $[Co(NO_2)_6]^{3-}$,$[Cu(NH_3)_4]^{2+}$,and $[Co(C_2O_4)_3]^{3-}$ are homoleptic complexes because they contain only one type of ligand.
In the complex $[Fe(H_2O)_5(NCS)]^{2+}$,the central metal ion $Fe^{2+}$ is bonded to two different types of ligands: $H_2O$ and $NCS^-$.
Therefore,it is a heteroleptic complex.
In contrast,the other complexes $[Co(NO_2)_6]^{3-}$,$[Cu(NH_3)_4]^{2+}$,and $[Co(C_2O_4)_3]^{3-}$ are homoleptic complexes because they contain only one type of ligand.
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229
ChemistryEasyMCQMHT CET · 2024
Identify a complex having a monodentate ligand from the following:
A
Tetracyanonickelate$(II)$ ion
B
Potassium trioxalatoaluminate$(III)$
C
Trioxalatocobaltate$(III)$ ion
D
bis(ethylenediamine)dithiocyanato platinum$(IV)$
Solution
(A) monodentate ligand is a ligand that binds to the central metal atom through only one donor atom.
$CN^{-}$ (cyanide) is a monodentate ligand.
$C_2O_4^{2-}$ (oxalate) is a bidentate ligand.
$en$ (ethylenediamine) is a bidentate ligand.
$SCN^{-}$ (thiocyanato) is an ambidentate ligand,which acts as a monodentate ligand.
In the given options,Tetracyanonickelate$(II)$ ion,$[Ni(CN)_4]^{2-}$,contains the monodentate ligand $CN^{-}$. Therefore,option $A$ is correct.
$CN^{-}$ (cyanide) is a monodentate ligand.
$C_2O_4^{2-}$ (oxalate) is a bidentate ligand.
$en$ (ethylenediamine) is a bidentate ligand.
$SCN^{-}$ (thiocyanato) is an ambidentate ligand,which acts as a monodentate ligand.
In the given options,Tetracyanonickelate$(II)$ ion,$[Ni(CN)_4]^{2-}$,contains the monodentate ligand $CN^{-}$. Therefore,option $A$ is correct.
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230
ChemistryDifficultMCQMHT CET · 2024
What is the number of moles of ionisable $Cl^{-}$ ions in a coordination complex if it forms $2 \ moles$ of $AgCl$ when treated with silver nitrate solution in excess?
A
$1$
B
$2$
C
$3$
D
$4$
Solution
(B) The reaction of a coordination complex with excess silver nitrate $(AgNO_3)$ precipitates the ionisable chloride ions as silver chloride $(AgCl)$.
According to the stoichiometry of the reaction:
$[M(X)_n]Cl_2 + 2AgNO_3 \rightarrow [M(X)_n](NO_3)_2 + 2AgCl$
Since $1 \ mole$ of the complex reacts to form $2 \ moles$ of $AgCl$,it indicates that there are $2 \ moles$ of ionisable $Cl^{-}$ ions present outside the coordination sphere.
According to the stoichiometry of the reaction:
$[M(X)_n]Cl_2 + 2AgNO_3 \rightarrow [M(X)_n](NO_3)_2 + 2AgCl$
Since $1 \ mole$ of the complex reacts to form $2 \ moles$ of $AgCl$,it indicates that there are $2 \ moles$ of ionisable $Cl^{-}$ ions present outside the coordination sphere.
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231
ChemistryMediumMCQMHT CET · 2024
Which among the following is a cationic complex?
A
$Na_4[Fe(CN)_6]$
B
$[Co(NH_3)_5Cl]SO_4$
C
$[Ni(CO)_4]$
D
$K_3[Fe(CN)_6]$
Solution
(B) complex is cationic if the coordination sphere carries a positive charge.
$Na_4[Fe(CN)_6]$ dissociates into $4Na^+$ and $[Fe(CN)_6]^{4-}$,which is an anionic complex.
$[Co(NH_3)_5Cl]SO_4$ dissociates into $[Co(NH_3)_5Cl]^{2+}$ and $SO_4^{2-}$,which is a cationic complex.
$[Ni(CO)_4]$ is a neutral complex as it has no charge.
$K_3[Fe(CN)_6]$ dissociates into $3K^+$ and $[Fe(CN)_6]^{3-}$,which is an anionic complex.
Therefore,the correct option is $B$.
$Na_4[Fe(CN)_6]$ dissociates into $4Na^+$ and $[Fe(CN)_6]^{4-}$,which is an anionic complex.
$[Co(NH_3)_5Cl]SO_4$ dissociates into $[Co(NH_3)_5Cl]^{2+}$ and $SO_4^{2-}$,which is a cationic complex.
$[Ni(CO)_4]$ is a neutral complex as it has no charge.
$K_3[Fe(CN)_6]$ dissociates into $3K^+$ and $[Fe(CN)_6]^{3-}$,which is an anionic complex.
Therefore,the correct option is $B$.
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232
ChemistryEasyMCQMHT CET · 2024
Which of the following complexes contains an ambidentate ligand?
A
Triamminetrinitrocobalt $(III)$
B
Pentacarbonyliron $(0)$
C
Tetraamminecopper $(II)$ ion
D
Trioxalatocobaltate $(III)$ ion
Solution
(A) An ambidentate ligand is a ligand that can coordinate to a central metal atom through two different donor atoms.
In the complex Triamminetrinitrocobalt $(III)$,the nitro group $(-NO_2)$ is an ambidentate ligand because it can coordinate through either the nitrogen atom (nitro) or the oxygen atom (nitrito).
In the complex Triamminetrinitrocobalt $(III)$,the nitro group $(-NO_2)$ is an ambidentate ligand because it can coordinate through either the nitrogen atom (nitro) or the oxygen atom (nitrito).
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233
ChemistryEasyMCQMHT CET · 2024
What is the $EAN$ of $Zn$ in $[Zn(NH_3)_4]^{2+}$?
A
$32$
B
$28$
C
$30$
D
$36$
Solution
(D) The atomic number $(Z)$ of $Zn$ is $30$.
In $[Zn(NH_3)_4]^{2+}$,the oxidation state of $Zn$ is $+2$,so $X = 2$.
The number of electrons donated by $4$ $NH_3$ ligands is $Y = 4 \times 2 = 8$.
The $EAN$ is calculated as $EAN = Z - X + Y$.
$EAN = 30 - 2 + 8 = 36$.
In $[Zn(NH_3)_4]^{2+}$,the oxidation state of $Zn$ is $+2$,so $X = 2$.
The number of electrons donated by $4$ $NH_3$ ligands is $Y = 4 \times 2 = 8$.
The $EAN$ is calculated as $EAN = Z - X + Y$.
$EAN = 30 - 2 + 8 = 36$.
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234
ChemistryEasyMCQMHT CET · 2024
Which of the following ligands is neutral?
A
Ammine
B
Chloro
C
Sulphato
D
Nitrito
Solution
(A) $NH_3$ is a neutral ligand.
All other options ($Chloro$,$Sulphato$,$Nitrito$) are anionic ligands.
Anionic ligands typically end with the suffix '$o$'.
All other options ($Chloro$,$Sulphato$,$Nitrito$) are anionic ligands.
Anionic ligands typically end with the suffix '$o$'.
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235
ChemistryEasyMCQMHT CET · 2024
What is the number of moles of donor atoms present in one mole of oxalate ion?
A
$1$
B
$2$
C
$6$
D
$4$
Solution
(B) The oxalate ion is represented by the formula $C_2O_4^{2-}$.
In the structure of the oxalate ion,there are two oxygen atoms that carry a negative charge and possess lone pairs of electrons,which act as donor atoms for coordination with a central metal ion.
Therefore,each oxalate ion has $2$ donor atoms.
Consequently,in $1 \text{ mole}$ of oxalate ion,there are $2 \text{ moles}$ of donor atoms.
In the structure of the oxalate ion,there are two oxygen atoms that carry a negative charge and possess lone pairs of electrons,which act as donor atoms for coordination with a central metal ion.
Therefore,each oxalate ion has $2$ donor atoms.
Consequently,in $1 \text{ mole}$ of oxalate ion,there are $2 \text{ moles}$ of donor atoms.
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236
ChemistryEasyMCQMHT CET · 2024
Identify the formula of pentaaqua-isothiocyanatoiron$(III)$ ion from the following.
A
$[Fe(H_2O)_5(SCN)]^{2+}$
B
$[Fe(H_2O)_5(NCS)]^{2+}$
C
$[Fe(H_2O)_3(NCS)_2]^{3+}$
D
$[Fe(H_2O)_2(SCN)_3]^{5+}$
Solution
(B) Pentaaqua means there are five water molecules,$(H_2O)_5$,ligated to the central metal.
Isothiocyanato refers to the presence of one isothiocyanate ligand,represented as $(NCS^-)$,which is $N$-bonded.
Iron$(III)$ ion indicates that the oxidation state of iron is $+3$.
The overall charge on the complex is calculated as: $\text{Charge} = \text{Oxidation state of } Fe + \text{Charge of } H_2O + \text{Charge of } NCS^- = (+3) + 0 + (-1) = +2$.
Thus,the formula is $[Fe(H_2O)_5(NCS)]^{2+}$.
Option $B$ is the correct answer.
Isothiocyanato refers to the presence of one isothiocyanate ligand,represented as $(NCS^-)$,which is $N$-bonded.
Iron$(III)$ ion indicates that the oxidation state of iron is $+3$.
The overall charge on the complex is calculated as: $\text{Charge} = \text{Oxidation state of } Fe + \text{Charge of } H_2O + \text{Charge of } NCS^- = (+3) + 0 + (-1) = +2$.
Thus,the formula is $[Fe(H_2O)_5(NCS)]^{2+}$.
Option $B$ is the correct answer.
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237
ChemistryMediumMCQMHT CET · 2024
What is the $IUPAC$ name of Ethylmethylisopropylamine?
A
$N$-Methyl-$N$-isopropylethanamine
B
$N$-Ethyl-$N$-methylpropan$-1-$amine
C
$N$-Ethyl-$N$-methylpropan$-2-$amine
D
$N$-Ethyl-$N$-isopropylmethanamine
Solution
(C) The given structure is $CH_3-CH(CH_3)-N(CH_3)(C_2H_5)$.
In $IUPAC$ nomenclature for secondary and tertiary amines,the longest carbon chain attached to the nitrogen atom is selected as the parent alkane.
Here,the longest chain is a $3$-carbon chain (propane) attached to the nitrogen at the $2$-position.
The other two alkyl groups attached to the nitrogen are ethyl and methyl groups,which are named as $N$-substituents.
Thus,the $IUPAC$ name is $N$-Ethyl-$N$-methylpropan$-2-$amine.
In $IUPAC$ nomenclature for secondary and tertiary amines,the longest carbon chain attached to the nitrogen atom is selected as the parent alkane.
Here,the longest chain is a $3$-carbon chain (propane) attached to the nitrogen at the $2$-position.
The other two alkyl groups attached to the nitrogen are ethyl and methyl groups,which are named as $N$-substituents.
Thus,the $IUPAC$ name is $N$-Ethyl-$N$-methylpropan$-2-$amine.
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238
ChemistryEasyMCQMHT CET · 2024
Which from the following is the formula of sodium hexafluoroaluminate$(III)$?
A
$Na_3[AlF_6]$
B
$Na[AlF_6]$
C
$Na_2[AlF_6]^{2+}$
D
$[Al(NaF)_6]^{2+}$
Solution
(A) The name sodium hexafluoroaluminate$(III)$ indicates the presence of sodium ions $(Na^+)$ and the complex anion hexafluoroaluminate$(III)$.
In the complex anion $[AlF_6]^{n-}$,the oxidation state of $Al$ is $+3$ and each $F$ is $-1$.
Therefore,the charge $n$ on the complex is calculated as: $x + 6(-1) = -3$,so $x = +3$.
The complex ion is $[AlF_6]^{3-}$.
To balance the charge of the complex ion,three $Na^+$ ions are required.
Thus,the formula is $Na_3[AlF_6]$.
In the complex anion $[AlF_6]^{n-}$,the oxidation state of $Al$ is $+3$ and each $F$ is $-1$.
Therefore,the charge $n$ on the complex is calculated as: $x + 6(-1) = -3$,so $x = +3$.
The complex ion is $[AlF_6]^{3-}$.
To balance the charge of the complex ion,three $Na^+$ ions are required.
Thus,the formula is $Na_3[AlF_6]$.
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239
ChemistryMediumMCQMHT CET · 2024
Which from the following ligands is able to form linkage isomers?
A
Aqua
B
Iodo
C
Ammine
D
Nitro
Solution
(D) Linkage isomerism is exhibited by ambidentate ligands.
An ambidentate ligand is a ligand that can coordinate to the central metal atom through two different donor atoms.
Among the given options,$Nitro$ $(-NO_2^-)$ is an ambidentate ligand because it can coordinate through either the nitrogen atom $(-NO_2)$ or the oxygen atom $(-ONO)$.
Other options like $Aqua$ $(H_2O)$,$Iodo$ $(I^-)$,and $Ammine$ $(NH_3)$ are monodentate ligands that do not exhibit linkage isomerism.
An ambidentate ligand is a ligand that can coordinate to the central metal atom through two different donor atoms.
Among the given options,$Nitro$ $(-NO_2^-)$ is an ambidentate ligand because it can coordinate through either the nitrogen atom $(-NO_2)$ or the oxygen atom $(-ONO)$.
Other options like $Aqua$ $(H_2O)$,$Iodo$ $(I^-)$,and $Ammine$ $(NH_3)$ are monodentate ligands that do not exhibit linkage isomerism.
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240
ChemistryMediumMCQMHT CET · 2024
What type of isomerism is exhibited by $[Cr(H_2O)_6]Cl_3$ and $[Cr(H_2O)_5Cl]Cl_2 \cdot H_2O$?
A
Coordinate
B
Solvate
C
Ionization
D
Linkage
Solution
(B) The complexes $[Cr(H_2O)_6]Cl_3$ and $[Cr(H_2O)_5Cl]Cl_2 \cdot H_2O$ differ in the number of water molecules directly coordinated to the central metal ion $Cr^{3+}$ versus those present as lattice water (solvent of crystallization).
This type of isomerism,where the solvent molecule (in this case,$H_2O$) acts as a ligand in one isomer and as a free solvent molecule in another,is known as solvate isomerism (also called hydrate isomerism).
Therefore,the correct answer is $B$.
This type of isomerism,where the solvent molecule (in this case,$H_2O$) acts as a ligand in one isomer and as a free solvent molecule in another,is known as solvate isomerism (also called hydrate isomerism).
Therefore,the correct answer is $B$.
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241
ChemistryEasyMCQMHT CET · 2024
What is the number of unpaired electrons present in $[CoF_6]^{3-}$?
A
$4$
B
$3$
C
$2$
D
$1$
Solution
(A) The atomic number of $Co$ is $27$. The electronic configuration of $Co$ is $[Ar] 3d^7 4s^2$.
In $[CoF_6]^{3-}$,the oxidation state of $Co$ is $x + 6(-1) = -3$,so $x = +3$.
The electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$.
$F^-$ is a weak field ligand,so it does not cause pairing of electrons in the $3d$ orbitals.
Therefore,the $3d^6$ configuration remains as $t_{2g}^4 e_g^2$,which corresponds to $4$ unpaired electrons.
In $[CoF_6]^{3-}$,the oxidation state of $Co$ is $x + 6(-1) = -3$,so $x = +3$.
The electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$.
$F^-$ is a weak field ligand,so it does not cause pairing of electrons in the $3d$ orbitals.
Therefore,the $3d^6$ configuration remains as $t_{2g}^4 e_g^2$,which corresponds to $4$ unpaired electrons.
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242
ChemistryEasyMCQMHT CET · 2024
Identify the type of hybridization present in $[Ni(CN)_4]^{2-}$.
A
$sp^3 d^2$
B
$dsp^2$
C
$sp^3$
D
$d^2 sp^3$
Solution
(B) In $[Ni(CN)_4]^{2-}$,the oxidation state of nickel is $+2$.
The valence shell electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$.
Since $CN^-$ is a strong field ligand,the $3d$ electrons are paired prior to hybridization.
This results in one vacant $3d$ orbital,one $4s$ orbital,and two $4p$ orbitals,which undergo $dsp^2$ hybridization to accommodate the four electron pairs donated by the four $CN^-$ ligands.
The valence shell electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$.
Since $CN^-$ is a strong field ligand,the $3d$ electrons are paired prior to hybridization.
This results in one vacant $3d$ orbital,one $4s$ orbital,and two $4p$ orbitals,which undergo $dsp^2$ hybridization to accommodate the four electron pairs donated by the four $CN^-$ ligands.
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243
ChemistryMediumMCQMHT CET · 2024
What type of hybridization is present in $[Co(NH_3)_6]^{3+}$ complex?
A
$d^2sp^3$
B
$sp^3$
C
$sp^3d^2$
D
$dsp^3$
Solution
(A) The atomic number of Cobalt $(Co)$ is $27$. The electronic configuration of $Co$ is $[Ar] 3d^7 4s^2$.
In the complex $[Co(NH_3)_6]^{3+}$,the oxidation state of $Co$ is $+3$. Thus,the configuration of $Co^{3+}$ is $[Ar] 3d^6$.
$NH_3$ is a strong field ligand,which causes the pairing of electrons in the $3d$ orbitals.
After pairing,two $3d$ orbitals,one $4s$ orbital,and three $4p$ orbitals become available for hybridization.
Therefore,the hybridization involved is $d^2sp^3$.
In the complex $[Co(NH_3)_6]^{3+}$,the oxidation state of $Co$ is $+3$. Thus,the configuration of $Co^{3+}$ is $[Ar] 3d^6$.
$NH_3$ is a strong field ligand,which causes the pairing of electrons in the $3d$ orbitals.
After pairing,two $3d$ orbitals,one $4s$ orbital,and three $4p$ orbitals become available for hybridization.
Therefore,the hybridization involved is $d^2sp^3$.
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244
ChemistryDifficultMCQMHT CET · 2024
Identify the total number of complexes having bidentate ligands in them from the following list of complexes:
$a$) Tetracyanonickelate$(II)$ ion
$b$) Trioxalatocobaltate$(III)$ ion
$c$) Sodium hexafluoroaluminate$(III)$
$d$) bis(ethylenediamine)dithiocyanatoplatinum$(IV)$
$a$) Tetracyanonickelate$(II)$ ion
$b$) Trioxalatocobaltate$(III)$ ion
$c$) Sodium hexafluoroaluminate$(III)$
$d$) bis(ethylenediamine)dithiocyanatoplatinum$(IV)$
A
$1$
B
$2$
C
$3$
D
$4$
Solution
(B) To identify the complexes with bidentate ligands,we analyze the ligands present in each complex:
$a$) Tetracyanonickelate$(II)$ ion: $[Ni(CN)_4]^{2-}$. The ligand is cyanide $(CN^-)$,which is a monodentate ligand.
$b$) Trioxalatocobaltate$(III)$ ion: $[Co(C_2O_4)_3]^{3-}$. The ligand is oxalate $(C_2O_4^{2-})$,which is a bidentate ligand.
$c$) Sodium hexafluoroaluminate$(III)$: $Na_3[AlF_6]$. The ligand is fluoride $(F^-)$,which is a monodentate ligand.
$d$) bis(ethylenediamine)dithiocyanatoplatinum$(IV)$: $[Pt(en)_2(SCN)_2]^{2+}$. The ligand ethylenediamine $(en)$ is a bidentate ligand.
Thus,complexes $(b)$ and $(d)$ contain bidentate ligands.
The total number of such complexes is $2$.
$a$) Tetracyanonickelate$(II)$ ion: $[Ni(CN)_4]^{2-}$. The ligand is cyanide $(CN^-)$,which is a monodentate ligand.
$b$) Trioxalatocobaltate$(III)$ ion: $[Co(C_2O_4)_3]^{3-}$. The ligand is oxalate $(C_2O_4^{2-})$,which is a bidentate ligand.
$c$) Sodium hexafluoroaluminate$(III)$: $Na_3[AlF_6]$. The ligand is fluoride $(F^-)$,which is a monodentate ligand.
$d$) bis(ethylenediamine)dithiocyanatoplatinum$(IV)$: $[Pt(en)_2(SCN)_2]^{2+}$. The ligand ethylenediamine $(en)$ is a bidentate ligand.
Thus,complexes $(b)$ and $(d)$ contain bidentate ligands.
The total number of such complexes is $2$.
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245
ChemistryDifficultMCQMHT CET · 2024
Which from the following statements about $[Co(NH_3)_6]^{3+}$ complex is $NOT$ correct?
A
Prior to hybridization $Co^{3+}$ possesses four unpaired electrons.
B
This complex has all electrons paired.
C
It is a high spin complex.
D
It is a diamagnetic complex.
Solution
(C) In $[Co(NH_3)_6]^{3+}$,the oxidation state of cobalt is $+3$. The valence shell electronic configuration of $Co^{3+}$ $(3d^6)$ is represented as: $3d$ ($4$ unpaired electrons),$4s$ (empty),$4p$ (empty).
Since $NH_3$ is a strong field ligand,it causes pairing of electrons in the $3d$ orbitals.
After pairing,the configuration becomes $t_{2g}^6 e_g^0$,meaning all electrons are paired.
This makes the complex diamagnetic and a low spin complex.
Therefore,the statement that it is a high spin complex is incorrect.
Since $NH_3$ is a strong field ligand,it causes pairing of electrons in the $3d$ orbitals.
After pairing,the configuration becomes $t_{2g}^6 e_g^0$,meaning all electrons are paired.
This makes the complex diamagnetic and a low spin complex.
Therefore,the statement that it is a high spin complex is incorrect.
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246
ChemistryMediumMCQMHT CET · 2024
Which of the following coordination complexes is a heteroleptic complex?
A
Pentacarbonyliron$(0)$
B
Triamminetrinitrocobalt$(III)$
C
Tetraamminecopper$(II)$ ion
D
Tetracyanonickelate$(II)$ ion
Solution
(B) heteroleptic complex is one in which the central metal atom or ion is bonded to more than one type of donor group or ligand.
In $[Co(NH_3)_3(NO_2)_3]$,the central metal ion $Co^{3+}$ is bonded to two different types of ligands: $NH_3$ and $NO_2^-$.
Therefore,it is a heteroleptic complex.
In $[Co(NH_3)_3(NO_2)_3]$,the central metal ion $Co^{3+}$ is bonded to two different types of ligands: $NH_3$ and $NO_2^-$.
Therefore,it is a heteroleptic complex.
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247
ChemistryMediumMCQMHT CET · 2024
What type of ligand is the oxalate ion?
A
Monodentate
B
Bidentate
C
Ambidentate
D
Hexadentate
Solution
(B) The oxalate ion,$(C_2O_4)^{2-}$,is a bidentate ligand.
It contains two negatively charged oxygen atoms that act as donor atoms to coordinate with the central metal ion.
Thus,it forms two coordinate bonds with the metal atom.
It contains two negatively charged oxygen atoms that act as donor atoms to coordinate with the central metal ion.
Thus,it forms two coordinate bonds with the metal atom.
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248
ChemistryMediumMCQMHT CET · 2024
Which among the following is a correct decreasing order of thermodynamic stability of the complexes?
A
$[Co(NH_3)_6]^{3+} > [Ag(CN)_2]^{-} > [Cu(CN)_4]^{2-}$
B
$[Co(NH_3)_6]^{3+} > [Cu(CN)_4]^{2-} > [Ag(CN)_2]^{-}$
C
$[Ag(CN)_2]^{-} > [Co(NH_3)_6]^{3+} > [Cu(CN)_4]^{2-}$
D
$[Cu(CN)_4]^{2-} > [Ag(CN)_2]^{-} > [Co(NH_3)_6]^{3+}$
Solution
(B) The thermodynamic stability of a coordination complex is primarily determined by the charge density of the central metal ion.
Higher charge on the central metal ion leads to stronger electrostatic attraction with the ligands,resulting in higher stability.
The oxidation states of the central metal ions in the given complexes are:
$Co$ in $[Co(NH_3)_6]^{3+}$ is $+3$.
$Cu$ in $[Cu(CN)_4]^{2-}$ is $+2$.
$Ag$ in $[Ag(CN)_2]^{-}$ is $+1$.
Therefore,the order of stability follows the order of oxidation states: $[Co(NH_3)_6]^{3+} > [Cu(CN)_4]^{2-} > [Ag(CN)_2]^{-}$.
Higher charge on the central metal ion leads to stronger electrostatic attraction with the ligands,resulting in higher stability.
The oxidation states of the central metal ions in the given complexes are:
$Co$ in $[Co(NH_3)_6]^{3+}$ is $+3$.
$Cu$ in $[Cu(CN)_4]^{2-}$ is $+2$.
$Ag$ in $[Ag(CN)_2]^{-}$ is $+1$.
Therefore,the order of stability follows the order of oxidation states: $[Co(NH_3)_6]^{3+} > [Cu(CN)_4]^{2-} > [Ag(CN)_2]^{-}$.
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249
ChemistryEasyMCQMHT CET · 2024
Which from the following is a correct stability order of complexes formed by metal ions if the ligand remains the same?
A
$Fe^{2+} > Mn^{2+} > Co^{2+}$
B
$Co^{2+} > Fe^{2+} > Mn^{2+}$
C
$Co^{2+} > Mn^{2+} > Fe^{2+}$
D
$Fe^{2+} > Co^{2+} > Mn^{2+}$
Solution
(B) The stability order of divalent metal ion complexes is governed by the Irving-Williams series.
According to this series,for the same ligand,the stability of complexes of divalent metal ions follows the order: $Mn^{2+} < Fe^{2+} < Co^{2+} < Ni^{2+} < Cu^{2+} > Zn^{2+}$.
This trend is primarily due to the decrease in ionic radius and the increase in crystal field stabilization energy $(CFSE)$ across the series.
Therefore,the correct stability order for the given ions is $Co^{2+} > Fe^{2+} > Mn^{2+}$.
According to this series,for the same ligand,the stability of complexes of divalent metal ions follows the order: $Mn^{2+} < Fe^{2+} < Co^{2+} < Ni^{2+} < Cu^{2+} > Zn^{2+}$.
This trend is primarily due to the decrease in ionic radius and the increase in crystal field stabilization energy $(CFSE)$ across the series.
Therefore,the correct stability order for the given ions is $Co^{2+} > Fe^{2+} > Mn^{2+}$.
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250
ChemistryEasyMCQMHT CET · 2024
What is the number of electrons lost by $Cr$ in a complex $[Cr(CO)_6]$?
A
$0$
B
$4$
C
$6$
D
$2$
Solution
(A) The complex is $[Cr(CO)_6]$.
Carbonyl $(CO)$ is a neutral ligand with an oxidation state of $0$.
Let the oxidation state of $Cr$ be $x$.
$x + 6(0) = 0$
$x = 0$.
Since the oxidation state of $Cr$ is $0$,the number of electrons lost by $Cr$ is $0$.
Carbonyl $(CO)$ is a neutral ligand with an oxidation state of $0$.
Let the oxidation state of $Cr$ be $x$.
$x + 6(0) = 0$
$x = 0$.
Since the oxidation state of $Cr$ is $0$,the number of electrons lost by $Cr$ is $0$.
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