ChemistryQ1–13 of 13 questions
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1
ChemistryEasyMCQGSEB · 2021
What is the electronic configuration of $Cr$?
A
$[Ar] 3d^4 4s^2$
B
$[Ar] 3d^5 4s^1$
C
$[Ar] 3d^5 4s^0$
D
$[Ar] 3d^4 4s^0$
Solution
(B) The atomic number of Chromium $(Cr)$ is $24$.
According to the Aufbau principle,the expected configuration is $[Ar] 3d^4 4s^2$.
However,half-filled $d$-orbitals are more stable due to exchange energy and symmetry.
Therefore,one electron from the $4s$ orbital shifts to the $3d$ orbital,resulting in the stable configuration $[Ar] 3d^5 4s^1$.
Thus,the correct option is $B$.
According to the Aufbau principle,the expected configuration is $[Ar] 3d^4 4s^2$.
However,half-filled $d$-orbitals are more stable due to exchange energy and symmetry.
Therefore,one electron from the $4s$ orbital shifts to the $3d$ orbital,resulting in the stable configuration $[Ar] 3d^5 4s^1$.
Thus,the correct option is $B$.
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2
ChemistryEasyMCQGSEB · 2021
Which of the following is the incorrect equation for the rate constant of a zero-order reaction?
A
$K = \frac{[R]_0 - [R]}{t}$
B
$[R] = -Kt + [R]_0$
C
$-K = \frac{[R] - [R]_0}{t}$
D
$[R]_0 + [R] = -Kt$
Solution
(D) For a zero-order reaction,the integrated rate equation is given by $[R] = -Kt + [R]_0$.
Rearranging this,we get $Kt = [R]_0 - [R]$,which implies $K = \frac{[R]_0 - [R]}{t}$.
Also,$-K = \frac{[R] - [R]_0}{t}$ is a valid rearrangement of the same equation.
However,the equation $[R]_0 + [R] = -Kt$ is mathematically incorrect as it does not represent the relationship between concentration and time for a zero-order reaction.
Therefore,option $D$ is the incorrect equation.
Rearranging this,we get $Kt = [R]_0 - [R]$,which implies $K = \frac{[R]_0 - [R]}{t}$.
Also,$-K = \frac{[R] - [R]_0}{t}$ is a valid rearrangement of the same equation.
However,the equation $[R]_0 + [R] = -Kt$ is mathematically incorrect as it does not represent the relationship between concentration and time for a zero-order reaction.
Therefore,option $D$ is the incorrect equation.
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3
ChemistryEasyMCQGSEB · 2021
$A$ first order reaction is found to have a rate constant,$k = 5.5 \times 10^{-14} \ s^{-1}$. The half life of reaction is . . . . . . .
A
$1.26 \times 10^{13} \ s$
B
$1.26 \times 10^{14} \ s$
C
$6.93 \times 10^{14} \ s$
D
$12.6 \times 10^{15} \ s$
Solution
(A) For a first order reaction,the half-life $(t_{1/2})$ is given by the formula: $t_{1/2} = \frac{0.693}{k}$.
Given $k = 5.5 \times 10^{-14} \ s^{-1}$.
Substituting the value of $k$ in the formula:
$t_{1/2} = \frac{0.693}{5.5 \times 10^{-14}} \ s$.
$t_{1/2} = 0.126 \times 10^{14} \ s$.
$t_{1/2} = 1.26 \times 10^{13} \ s$.
Given $k = 5.5 \times 10^{-14} \ s^{-1}$.
Substituting the value of $k$ in the formula:
$t_{1/2} = \frac{0.693}{5.5 \times 10^{-14}} \ s$.
$t_{1/2} = 0.126 \times 10^{14} \ s$.
$t_{1/2} = 1.26 \times 10^{13} \ s$.
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4
ChemistryEasyMCQGSEB · 2021
Which of the following is not a chelating ligand?
A
Oxalato
B
$NH_3$
C
$EDTA$
D
Ethane-$1, 2$-diamine
Solution
(B) chelating ligand is a polydentate ligand that can form a ring structure with a central metal atom.
$NH_3$ (ammine) is a monodentate ligand,meaning it binds to the central metal atom through only one donor atom.
Therefore,$NH_3$ cannot form a ring structure and is not a chelating ligand.
Oxalato,$EDTA$,and Ethane-$1, 2$-diamine are all polydentate ligands capable of chelation.
$NH_3$ (ammine) is a monodentate ligand,meaning it binds to the central metal atom through only one donor atom.
Therefore,$NH_3$ cannot form a ring structure and is not a chelating ligand.
Oxalato,$EDTA$,and Ethane-$1, 2$-diamine are all polydentate ligands capable of chelation.
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5
ChemistryEasyMCQGSEB · 2021
Which of the following compounds does not show optical activity?
A
$Cis-[CrCl_2(ox)_2]^{3-}$
B
$Cis-[CoBr_2(en)_2]^{+}$
C
$Cis-[Fe(NH_3)_2(CN)_4]^{-}$
D
$Cis-[PtCl_2(en)_2]^{2+}$
Solution
(C) Optical activity in coordination compounds requires the absence of a plane of symmetry or a center of inversion.
$Cis-[CrCl_2(ox)_2]^{3-}$ has no plane of symmetry and is optically active.
$Cis-[CoBr_2(en)_2]^{+}$ has no plane of symmetry and is optically active.
$Cis-[Fe(NH_3)_2(CN)_4]^{-}$ possesses a plane of symmetry (the plane containing the $Fe$,$2NH_3$,and $2CN$ ligands),making it achiral and optically inactive.
$Cis-[PtCl_2(en)_2]^{2+}$ is optically active due to the lack of a plane of symmetry.
Therefore,the correct option is $C$.
$Cis-[CrCl_2(ox)_2]^{3-}$ has no plane of symmetry and is optically active.
$Cis-[CoBr_2(en)_2]^{+}$ has no plane of symmetry and is optically active.
$Cis-[Fe(NH_3)_2(CN)_4]^{-}$ possesses a plane of symmetry (the plane containing the $Fe$,$2NH_3$,and $2CN$ ligands),making it achiral and optically inactive.
$Cis-[PtCl_2(en)_2]^{2+}$ is optically active due to the lack of a plane of symmetry.
Therefore,the correct option is $C$.
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6
ChemistryEasyMCQGSEB · 2021
$Cu^{2+}$ aqueous solution has . . . . . . colour.
A
Blue
B
Violet
C
Green
D
Yellow
Solution
(A) The $Cu^{2+}$ ion in aqueous solution exists as the hydrated complex $[Cu(H_2O)_6]^{2+}$.
Due to the presence of one unpaired electron in the $d$-orbital,$d-d$ transitions occur when light is absorbed.
The absorption of energy from the visible region results in the transmission of the complementary colour,which is blue.
Due to the presence of one unpaired electron in the $d$-orbital,$d-d$ transitions occur when light is absorbed.
The absorption of energy from the visible region results in the transmission of the complementary colour,which is blue.
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7
ChemistryEasyMCQGSEB · 2021
If a current of $0.5 \ A$ flows through a metallic wire for $2 \ hours$,then how many coulombs would have flown through the wire (in $C$)?
A
$360$
B
$3000$
C
$36000$
D
$3600$
Solution
(D) The formula for charge $(Q)$ is given by $Q = I \times t$,where $I$ is the current in amperes and $t$ is the time in seconds.
Given: $I = 0.5 \ A$,$t = 2 \ hours = 2 \times 3600 \ s = 7200 \ s$.
Calculation: $Q = 0.5 \ A \times 7200 \ s = 3600 \ C$.
Therefore,the total charge that flows through the wire is $3600 \ C$.
Given: $I = 0.5 \ A$,$t = 2 \ hours = 2 \times 3600 \ s = 7200 \ s$.
Calculation: $Q = 0.5 \ A \times 7200 \ s = 3600 \ C$.
Therefore,the total charge that flows through the wire is $3600 \ C$.
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8
ChemistryEasyMCQGSEB · 2021
In the following reaction,what is the value of equilibrium constant?
$Cu_{(s)} + 2 Ag_{(aq)}^{+} \rightarrow Cu_{(aq)}^{2+} + 2 Ag_{(s)}$
$E_{cell}^0 = 0.46 \ V$
$Cu_{(s)} + 2 Ag_{(aq)}^{+} \rightarrow Cu_{(aq)}^{2+} + 2 Ag_{(s)}$
$E_{cell}^0 = 0.46 \ V$
A
$3.92 \times 10^{10}$
B
$3.92 \times 10^{15}$
C
$39.2 \times 10^{15}$
D
$3.92 \times 10^{14}$
Solution
(B) The relationship between standard cell potential and equilibrium constant is given by the formula:
$E_{cell}^0 = \frac{0.0591}{n} \log K_c$
Here,$n = 2$ (number of electrons transferred).
$0.46 = \frac{0.0591}{2} \log K_c$
$\log K_c = \frac{0.46 \times 2}{0.0591} = 15.5668$
$K_c = \text{antilog}(15.5668) \approx 3.92 \times 10^{15}$
$E_{cell}^0 = \frac{0.0591}{n} \log K_c$
Here,$n = 2$ (number of electrons transferred).
$0.46 = \frac{0.0591}{2} \log K_c$
$\log K_c = \frac{0.46 \times 2}{0.0591} = 15.5668$
$K_c = \text{antilog}(15.5668) \approx 3.92 \times 10^{15}$
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9
ChemistryEasyMCQGSEB · 2021
For the following cell,the standard electrode potential $E_{cell}^0$ is . . . . . . .
$E_{Zn^{2+}/Zn}^0 = -0.76 \ V, E_{Cu^{2+}/Cu}^0 = 0.34 \ V$
$Zn | Zn^{2+} || Cu^{2+} | Cu$ (in $V$)
$E_{Zn^{2+}/Zn}^0 = -0.76 \ V, E_{Cu^{2+}/Cu}^0 = 0.34 \ V$
$Zn | Zn^{2+} || Cu^{2+} | Cu$ (in $V$)
A
$0.34$
B
$-0.76$
C
$1.10$
D
$-0.42$
Solution
(C) The standard cell potential $E_{cell}^0$ is calculated using the formula: $E_{cell}^0 = E_{cathode}^0 - E_{anode}^0$.
In the given cell $Zn | Zn^{2+} || Cu^{2+} | Cu$,$Zn$ acts as the anode and $Cu$ acts as the cathode.
Given: $E_{Cu^{2+}/Cu}^0 = 0.34 \ V$ and $E_{Zn^{2+}/Zn}^0 = -0.76 \ V$.
Substituting the values: $E_{cell}^0 = 0.34 \ V - (-0.76 \ V) = 0.34 \ V + 0.76 \ V = 1.10 \ V$.
In the given cell $Zn | Zn^{2+} || Cu^{2+} | Cu$,$Zn$ acts as the anode and $Cu$ acts as the cathode.
Given: $E_{Cu^{2+}/Cu}^0 = 0.34 \ V$ and $E_{Zn^{2+}/Zn}^0 = -0.76 \ V$.
Substituting the values: $E_{cell}^0 = 0.34 \ V - (-0.76 \ V) = 0.34 \ V + 0.76 \ V = 1.10 \ V$.
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10
ChemistryEasyMCQGSEB · 2021
If $5 \ g$ $NaOH$ is dissolved in $450 \ mL$ of solution,the molarity of the solution is: $(Na=23, O=16, H=1 \ g \ mol^{-1})$ (in $M$)
A
$27.8$
B
$0.278$
C
$2.78$
D
$278$
Solution
(B) The molar mass of $NaOH$ is $23 + 16 + 1 = 40 \ g \ mol^{-1}$.
Number of moles of $NaOH$ = $\frac{\text{mass}}{\text{molar mass}} = \frac{5 \ g}{40 \ g \ mol^{-1}} = 0.125 \ mol$.
Volume of solution = $450 \ mL = 0.450 \ L$.
Molarity $(M)$ = $\frac{\text{moles of solute}}{\text{volume of solution in } L} = \frac{0.125 \ mol}{0.450 \ L} \approx 0.278 \ M$.
Therefore,the correct option is $B$.
Number of moles of $NaOH$ = $\frac{\text{mass}}{\text{molar mass}} = \frac{5 \ g}{40 \ g \ mol^{-1}} = 0.125 \ mol$.
Volume of solution = $450 \ mL = 0.450 \ L$.
Molarity $(M)$ = $\frac{\text{moles of solute}}{\text{volume of solution in } L} = \frac{0.125 \ mol}{0.450 \ L} \approx 0.278 \ M$.
Therefore,the correct option is $B$.
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11
ChemistryEasyMCQGSEB · 2021
The pressure required for the reverse osmosis is quite high. For this,which of the following membranes is used?
A
Cellophane
B
Parchment
C
Cellulose acetate
D
Pig's bladder
Solution
(C) In the process of reverse osmosis,a semi-permeable membrane is required that can withstand high pressure. $Cellulose \ acetate$ is the material commonly used to prepare these semi-permeable membranes because it is permeable to water but impermeable to impurities and ions. Therefore,the correct option is $C$.
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12
ChemistryEasyMCQGSEB · 2021
$1.00 \text{ g}$ of a non-electrolyte solute dissolved in $50 \text{ g}$ of benzene lowered the freezing point of benzene by $0.40 \text{ K}$. The freezing point depression constant of benzene is $5.12 \text{ K kg mol}^{-1}$. Find the molar mass of the solute.
A
$280 \text{ g mol}^{-1}$
B
$356 \text{ g mol}^{-1}$
C
$562 \text{ g mol}^{-1}$
D
$256 \text{ g mol}^{-1}$
Solution
(D) Given: $w_2 = 1.00 \text{ g}$,$w_1 = 50 \text{ g}$,$\Delta T_f = 0.40 \text{ K}$,$K_f = 5.12 \text{ K kg mol}^{-1}$.
Using the formula: $\Delta T_f = \frac{K_f \times w_2 \times 1000}{M_2 \times w_1}$
Rearranging for $M_2$: $M_2 = \frac{K_f \times w_2 \times 1000}{\Delta T_f \times w_1}$
Substituting the values: $M_2 = \frac{5.12 \times 1.00 \times 1000}{0.40 \times 50}$
$M_2 = \frac{5120}{20} = 256 \text{ g mol}^{-1}$.
Using the formula: $\Delta T_f = \frac{K_f \times w_2 \times 1000}{M_2 \times w_1}$
Rearranging for $M_2$: $M_2 = \frac{K_f \times w_2 \times 1000}{\Delta T_f \times w_1}$
Substituting the values: $M_2 = \frac{5.12 \times 1.00 \times 1000}{0.40 \times 50}$
$M_2 = \frac{5120}{20} = 256 \text{ g mol}^{-1}$.
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13
ChemistryEasyMCQGSEB · 2021
In which solution,solute is liquid and solvent is gas?
A
Chloroform mixed with nitrogen gas
B
Ethanol dissolved in water
C
Camphor in nitrogen gas
D
Solution of hydrogen in palladium
Solution
(A) solution where the solute is a liquid and the solvent is a gas is known as a liquid-in-gas solution.
Chloroform $(CHCl_3)$ is a liquid at room temperature,and nitrogen gas $(N_2)$ acts as the solvent.
Therefore,chloroform mixed with nitrogen gas is an example of a liquid-in-gas solution.
Chloroform $(CHCl_3)$ is a liquid at room temperature,and nitrogen gas $(N_2)$ acts as the solvent.
Therefore,chloroform mixed with nitrogen gas is an example of a liquid-in-gas solution.
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