ChemistryQ1–55 of 55 questions
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1
ChemistryMCQAIPMT · 1996
$A$ point charge $+q$ is placed at the centre of a cube of side $L$. The electric flux emerging from the cube is
A
$\frac{q}{\varepsilon_0}$
B
Zero
C
$\frac{6qL^2}{\varepsilon_0}$
D
$\frac{q}{6L^2\varepsilon_0}$
Solution
(A) According to Gauss's Law,the total electric flux $\phi$ through any closed surface is equal to the net charge $q_{enc}$ enclosed by the surface divided by the permittivity of free space $\varepsilon_0$.
Mathematically,$\phi = \frac{q_{enc}}{\varepsilon_0}$.
In this problem,the point charge $+q$ is enclosed by the cube.
Therefore,the total electric flux emerging from the cube is $\phi = \frac{q}{\varepsilon_0}$.
Mathematically,$\phi = \frac{q_{enc}}{\varepsilon_0}$.
In this problem,the point charge $+q$ is enclosed by the cube.
Therefore,the total electric flux emerging from the cube is $\phi = \frac{q}{\varepsilon_0}$.
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2
ChemistryMCQAIPMT · 1996
If a convex lens of focal length $80 \, cm$ and a concave lens of focal length $50 \, cm$ are combined together, what will be their resulting power?
A
$+ 6.5 \, D$
B
$-6.5 \, D$
C
$+ 7.5 \, D$
D
$-0.75 \, D$
Solution
(D) The power of a lens is given by $P = \frac{1}{f(\text{in meters})}$.
For a convex lens, $f_1 = +80 \, cm = +0.8 \, m$. So, $P_1 = \frac{1}{0.8} = +1.25 \, D$.
For a concave lens, $f_2 = -50 \, cm = -0.5 \, m$. So, $P_2 = \frac{1}{-0.5} = -2.0 \, D$.
The resulting power of the combination is $P = P_1 + P_2$.
$P = 1.25 \, D + (-2.0 \, D) = -0.75 \, D$.
Therefore, the correct option is $D$.
For a convex lens, $f_1 = +80 \, cm = +0.8 \, m$. So, $P_1 = \frac{1}{0.8} = +1.25 \, D$.
For a concave lens, $f_2 = -50 \, cm = -0.5 \, m$. So, $P_2 = \frac{1}{-0.5} = -2.0 \, D$.
The resulting power of the combination is $P = P_1 + P_2$.
$P = 1.25 \, D + (-2.0 \, D) = -0.75 \, D$.
Therefore, the correct option is $D$.
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3
ChemistryMediumMCQAIPMT · 1996
$KMnO_4$ reacts with oxalic acid according to the equation,$2MnO_4^- + 5C_2O_4^{2-} + 16H^{+} \to 2Mn^{2+} + 10CO_2 + 8H_2O$. Here,$20 \ mL$ of $0.1 \ M$ $KMnO_4$ is equivalent to:
A
$20 \ mL$ of $0.5 \ M$ $H_2C_2O_4$
B
$50 \ mL$ of $0.1 \ M$ $H_2C_2O_4$
C
$50 \ mL$ of $0.5 \ M$ $H_2C_2O_4$
D
$20 \ mL$ of $0.1 \ M$ $H_2C_2O_4$
Solution
(B) From the balanced chemical equation: $2 \ mol$ of $MnO_4^-$ reacts with $5 \ mol$ of $C_2O_4^{2-}$.
Using the stoichiometry relation: $\frac{n_{MnO_4^-}}{2} = \frac{n_{C_2O_4^{2-}}}{5}$.
Given: $V_{MnO_4^-} = 20 \ mL$,$M_{MnO_4^-} = 0.1 \ M$.
$n_{MnO_4^-} = M \times V = 0.1 \times 20 = 2 \ mmol$.
Therefore,$n_{C_2O_4^{2-}} = \frac{5}{2} \times n_{MnO_4^-} = 2.5 \times 2 = 5 \ mmol$.
Checking option $(B)$: $50 \ mL$ of $0.1 \ M$ $H_2C_2O_4$ gives $n = 50 \times 0.1 = 5 \ mmol$. This matches the requirement.
Using the stoichiometry relation: $\frac{n_{MnO_4^-}}{2} = \frac{n_{C_2O_4^{2-}}}{5}$.
Given: $V_{MnO_4^-} = 20 \ mL$,$M_{MnO_4^-} = 0.1 \ M$.
$n_{MnO_4^-} = M \times V = 0.1 \times 20 = 2 \ mmol$.
Therefore,$n_{C_2O_4^{2-}} = \frac{5}{2} \times n_{MnO_4^-} = 2.5 \times 2 = 5 \ mmol$.
Checking option $(B)$: $50 \ mL$ of $0.1 \ M$ $H_2C_2O_4$ gives $n = 50 \times 0.1 = 5 \ mmol$. This matches the requirement.
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4
ChemistryMediumMCQAIPMT · 1996
In Bohr's model of an atom,when an electron jumps from $n = 1$ to $n = 3$,how much energy is absorbed?
A
$2.15 \times 10^{-11} \ erg$
B
$0.1911 \times 10^{-10} \ erg$
C
$2.389 \times 10^{-12} \ erg$
D
$0.239 \times 10^{-10} \ erg$
Solution
(B) According to Bohr's model,the energy of an electron in the $n^{th}$ orbit is given by $E_n = -\frac{2.179 \times 10^{-11}}{n^2} \ erg$.
The energy change for a transition from $n_1 = 1$ to $n_2 = 3$ is $\Delta E = E_3 - E_1$.
$\Delta E = -\frac{2.179 \times 10^{-11}}{3^2} - (-\frac{2.179 \times 10^{-11}}{1^2})$.
$\Delta E = 2.179 \times 10^{-11} \times (1 - \frac{1}{9}) = 2.179 \times 10^{-11} \times \frac{8}{9}$.
$\Delta E = 1.9368 \times 10^{-11} \ erg = 0.19368 \times 10^{-10} \ erg$.
Comparing with the given options,the closest value is $0.1911 \times 10^{-10} \ erg$ (Option $B$).
Since the electron moves from a lower energy level to a higher energy level,energy is absorbed.
The energy change for a transition from $n_1 = 1$ to $n_2 = 3$ is $\Delta E = E_3 - E_1$.
$\Delta E = -\frac{2.179 \times 10^{-11}}{3^2} - (-\frac{2.179 \times 10^{-11}}{1^2})$.
$\Delta E = 2.179 \times 10^{-11} \times (1 - \frac{1}{9}) = 2.179 \times 10^{-11} \times \frac{8}{9}$.
$\Delta E = 1.9368 \times 10^{-11} \ erg = 0.19368 \times 10^{-10} \ erg$.
Comparing with the given options,the closest value is $0.1911 \times 10^{-10} \ erg$ (Option $B$).
Since the electron moves from a lower energy level to a higher energy level,energy is absorbed.
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5
ChemistryMediumMCQAIPMT · 1996
Which one has minimum (nearly zero) dipole moment?
A
but-$1$-ene
B
$cis$-but-$2$-ene
C
$trans$-but-$2$-ene
D
$2$-methylprop-$1$-ene
Solution
(C) The dipole moment of a molecule depends on its geometry and the polarity of its bonds.
In $trans$-but-$2$-ene,the two methyl groups are on opposite sides of the double bond.
Due to the symmetry of the molecule,the bond dipoles of the $C-CH_3$ bonds cancel each other out.
Therefore,the net dipole moment of $trans$-but-$2$-ene is zero.
In $trans$-but-$2$-ene,the two methyl groups are on opposite sides of the double bond.
Due to the symmetry of the molecule,the bond dipoles of the $C-CH_3$ bonds cancel each other out.
Therefore,the net dipole moment of $trans$-but-$2$-ene is zero.
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6
ChemistryMediumMCQAIPMT · 1996
The structure and hybridisation of $Si(CH_3)_4$ is
A
Bent,$sp$
B
Trigonal,$sp^2$
C
Octahedral,$sp^3d$
D
Tetrahedral,$sp^3$
Solution
(D) In $Si(CH_3)_4$,the central silicon atom is bonded to $4$ methyl groups.
Since there are $4$ bond pairs and $0$ lone pairs around the central $Si$ atom,the steric number is $4+0 = 4$.
$A$ steric number of $4$ corresponds to $sp^3$ hybridisation and a tetrahedral geometry.
Since there are $4$ bond pairs and $0$ lone pairs around the central $Si$ atom,the steric number is $4+0 = 4$.
$A$ steric number of $4$ corresponds to $sp^3$ hybridisation and a tetrahedral geometry.
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7
ChemistryMediumMCQAIPMT · 1996
Which of the following mixtures of gases does not obey Dalton's law of partial pressure?
A
$O_2$ and $CO_2$
B
$N_2$ and $O_2$
C
$Cl_2$ and $O_2$
D
$NH_3$ and $HCl$
Solution
(D) Because $NH_3$ and $HCl$ gases react to form solid $NH_4Cl$ $(NH_3(g) + HCl(g) \rightarrow NH_4Cl(s))$.
Dalton's law of partial pressure is only applicable to mixtures of non-reacting gases.
Dalton's law of partial pressure is only applicable to mixtures of non-reacting gases.
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8
ChemistryMediumMCQAIPMT · 1996
At what temperature,the rate of effusion of $N_2$ would be $1.625$ times that of $SO_2$ at $50\,^{\circ}C$ ............ $K$
A
$110$
B
$173$
C
$373$
D
$273$
Solution
(C) According to Graham's Law of effusion,the rate of effusion $r$ is inversely proportional to the square root of the molar mass $M$ and directly proportional to the square root of the temperature $T$ for a given pressure.
$\frac{r_{N_2}}{r_{SO_2}} = \sqrt{\frac{M_{SO_2}}{M_{N_2}} \times \frac{T_{N_2}}{T_{SO_2}}}$
Given: $r_{N_2} = 1.625 \times r_{SO_2}$,$T_{SO_2} = 50 + 273 = 323\,K$,$M_{N_2} = 28\,g/mol$,$M_{SO_2} = 64\,g/mol$.
Substituting the values:
$1.625 = \sqrt{\frac{64}{28} \times \frac{T_{N_2}}{323}}$
Squaring both sides:
$(1.625)^2 = \frac{64}{28} \times \frac{T_{N_2}}{323}$
$2.640625 = 2.2857 \times \frac{T_{N_2}}{323}$
$T_{N_2} = \frac{2.640625 \times 323}{2.2857} \approx 373\,K$.
$\frac{r_{N_2}}{r_{SO_2}} = \sqrt{\frac{M_{SO_2}}{M_{N_2}} \times \frac{T_{N_2}}{T_{SO_2}}}$
Given: $r_{N_2} = 1.625 \times r_{SO_2}$,$T_{SO_2} = 50 + 273 = 323\,K$,$M_{N_2} = 28\,g/mol$,$M_{SO_2} = 64\,g/mol$.
Substituting the values:
$1.625 = \sqrt{\frac{64}{28} \times \frac{T_{N_2}}{323}}$
Squaring both sides:
$(1.625)^2 = \frac{64}{28} \times \frac{T_{N_2}}{323}$
$2.640625 = 2.2857 \times \frac{T_{N_2}}{323}$
$T_{N_2} = \frac{2.640625 \times 323}{2.2857} \approx 373\,K$.
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9
ChemistryMediumMCQAIPMT · 1996
The average kinetic energy of an ideal gas per molecule in $SI$ units at $25\,^oC$ will be
A
$6.17 \times 10^{-21} \text{ kJ}$
B
$6.17 \times 10^{-21} \text{ J}$
C
$6.17 \times 10^{-20} \text{ J}$
D
$6.17 \times 10^{-22} \text{ J}$
Solution
(B) The average kinetic energy per molecule of an ideal gas is given by the formula: $E_k = \frac{3}{2}kT$.
Here,$k$ is the Boltzmann constant $(1.38 \times 10^{-23} \text{ J/K})$ and $T$ is the temperature in Kelvin.
Given $T = 25\,^oC = 25 + 273 = 298 \text{ K}$.
Substituting the values: $E_k = \frac{3}{2} \times 1.38 \times 10^{-23} \times 298 \approx 6.17 \times 10^{-21} \text{ J}$.
Here,$k$ is the Boltzmann constant $(1.38 \times 10^{-23} \text{ J/K})$ and $T$ is the temperature in Kelvin.
Given $T = 25\,^oC = 25 + 273 = 298 \text{ K}$.
Substituting the values: $E_k = \frac{3}{2} \times 1.38 \times 10^{-23} \times 298 \approx 6.17 \times 10^{-21} \text{ J}$.
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10
ChemistryDifficultMCQAIPMT · 1996
The equilibrium constant for the reaction $N_2 + 3H_2 \rightleftharpoons 2NH_3$ is $K.$ Then,the equilibrium constant for the equilibrium $NH_3 \rightleftharpoons \frac{1}{2}N_2 + \frac{3}{2}H_2$ is
A
$1/K$
B
$1/K^2$
C
$\sqrt{K}$
D
$1/\sqrt{K}$
Solution
(D) For the reaction $N_2 + 3H_2 \rightleftharpoons 2NH_3$,the equilibrium constant is $K = \frac{[NH_3]^2}{[N_2][H_2]^3}$.
For the reaction $NH_3 \rightleftharpoons \frac{1}{2}N_2 + \frac{3}{2}H_2$,the equilibrium constant $K'$ is given by $K' = \frac{[N_2]^{1/2}[H_2]^{3/2}}{[NH_3]}$.
Comparing the two expressions,we see that $K' = \sqrt{\frac{1}{K}} = \frac{1}{\sqrt{K}}$.
For the reaction $NH_3 \rightleftharpoons \frac{1}{2}N_2 + \frac{3}{2}H_2$,the equilibrium constant $K'$ is given by $K' = \frac{[N_2]^{1/2}[H_2]^{3/2}}{[NH_3]}$.
Comparing the two expressions,we see that $K' = \sqrt{\frac{1}{K}} = \frac{1}{\sqrt{K}}$.
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11
ChemistryEasyMCQAIPMT · 1996
Which of the following is not a Lewis acid?
A
$BF_3$
B
$FeCl_3$
C
$SiF_4$
D
$C_2H_4$
Solution
(D) Lewis acid is defined as an electron pair acceptor.
$BF_3$ has an incomplete octet on the boron atom,making it an electron-deficient species and a Lewis acid.
$FeCl_3$ has an empty $d$-orbital and an incomplete octet,acting as a Lewis acid.
$SiF_4$ has vacant $d$-orbitals on the silicon atom,allowing it to accept electron pairs,thus acting as a Lewis acid.
$C_2H_4$ (ethene) has a $\pi$-bond but does not have an empty orbital to accept an electron pair in the standard sense of a Lewis acid.
Therefore,$C_2H_4$ is not a Lewis acid.
$BF_3$ has an incomplete octet on the boron atom,making it an electron-deficient species and a Lewis acid.
$FeCl_3$ has an empty $d$-orbital and an incomplete octet,acting as a Lewis acid.
$SiF_4$ has vacant $d$-orbitals on the silicon atom,allowing it to accept electron pairs,thus acting as a Lewis acid.
$C_2H_4$ (ethene) has a $\pi$-bond but does not have an empty orbital to accept an electron pair in the standard sense of a Lewis acid.
Therefore,$C_2H_4$ is not a Lewis acid.
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12
ChemistryMediumMCQAIPMT · 1996
Boron halides behave as Lewis acids because of their
A
Ionic nature
B
Acidic nature
C
Covalent nature
D
Electron deficient nature
Solution
(D) $Boron$ halides $(BX_3)$ have only $6$ electrons in the valence shell of the central $Boron$ atom.
Due to this electron deficiency,they have a strong tendency to accept a lone pair of electrons from a donor to complete their octet,which is the definition of a $Lewis$ acid.
For example,in $BF_3$,the $Boron$ atom is electron-deficient.
Due to this electron deficiency,they have a strong tendency to accept a lone pair of electrons from a donor to complete their octet,which is the definition of a $Lewis$ acid.
For example,in $BF_3$,the $Boron$ atom is electron-deficient.
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13
ChemistryDifficultMCQAIPMT · 1996
The ionic product of water at $25\,^{\circ}C$ is $10^{-14}$. The ionic product at $90\,^{\circ}C$ will be
A
$1 \times 10^{-20}$
B
$1 \times 10^{-12}$
C
$1 \times 10^{-14}$
D
$1 \times 10^{-16}$
Solution
(B) The auto-ionization of water is an endothermic process represented by the equation: $H_2O(l) \rightleftharpoons H^+(aq) + OH^-(aq)$.
According to Le Chatelier's principle,for an endothermic reaction,an increase in temperature shifts the equilibrium to the right.
Therefore,the concentration of $H^+$ and $OH^-$ ions increases,which leads to an increase in the ionic product of water $(K_w)$.
Since $90\,^{\circ}C > 25\,^{\circ}C$,the value of $K_w$ at $90\,^{\circ}C$ must be greater than $10^{-14}$.
Among the given options,only $1 \times 10^{-12}$ is greater than $10^{-14}$.
According to Le Chatelier's principle,for an endothermic reaction,an increase in temperature shifts the equilibrium to the right.
Therefore,the concentration of $H^+$ and $OH^-$ ions increases,which leads to an increase in the ionic product of water $(K_w)$.
Since $90\,^{\circ}C > 25\,^{\circ}C$,the value of $K_w$ at $90\,^{\circ}C$ must be greater than $10^{-14}$.
Among the given options,only $1 \times 10^{-12}$ is greater than $10^{-14}$.
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14
ChemistryMediumMCQAIPMT · 1996
The $pH$ value of a $\frac{N}{10} \ NaOH$ solution is:
A
$10$
B
$11$
C
$12$
D
$13$
Solution
(D) For a $\frac{N}{10} \ NaOH$ solution,the concentration of $OH^-$ ions is $[OH^-] = 10^{-1} \ M$.
Since $pOH = -\log[OH^-]$,we get $pOH = -\log(10^{-1}) = 1$.
Using the relation $pH + pOH = 14$ at $25^{\circ}C$,we find $pH = 14 - 1 = 13$.
Since $pOH = -\log[OH^-]$,we get $pOH = -\log(10^{-1}) = 1$.
Using the relation $pH + pOH = 14$ at $25^{\circ}C$,we find $pH = 14 - 1 = 13$.
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15
ChemistryMediumMCQAIPMT · 1996
Total number of moles for the reaction $2HI \rightleftharpoons H_2 + I_2$ at equilibrium,if $\alpha$ is the degree of dissociation,is:
A
$1$
B
$2 - \alpha$
C
$2$
D
$1 - \alpha$
Solution
(C) For the reaction: $2HI \rightleftharpoons H_2 + I_2$
Initial moles: $2 \text{ moles of } HI$,$0 \text{ moles of } H_2$,$0 \text{ moles of } I_2$.
At equilibrium: Moles of $HI = 2 - 2\alpha$,Moles of $H_2 = \alpha$,Moles of $I_2 = \alpha$.
Total moles at equilibrium $= (2 - 2\alpha) + \alpha + \alpha = 2$.
Initial moles: $2 \text{ moles of } HI$,$0 \text{ moles of } H_2$,$0 \text{ moles of } I_2$.
At equilibrium: Moles of $HI = 2 - 2\alpha$,Moles of $H_2 = \alpha$,Moles of $I_2 = \alpha$.
Total moles at equilibrium $= (2 - 2\alpha) + \alpha + \alpha = 2$.
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16
ChemistryMediumMCQAIPMT · 1996
Which of the following is the correct mathematical expression for the first law of thermodynamics?
A
$\Delta U = q + w$
B
$\Delta W = \Delta U + \Delta Q$
C
$\Delta U = \Delta W + \Delta Q$
D
None of these
Solution
(A) The first law of thermodynamics states that the change in internal energy $(\Delta U)$ of a system is equal to the sum of heat added to the system $(q)$ and the work done on the system $(w)$.
Mathematically,this is expressed as $\Delta U = q + w$.
Mathematically,this is expressed as $\Delta U = q + w$.
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17
ChemistryEasyMCQAIPMT · 1996
Which of the following has the smallest size?
A
$Na^{+}$
B
$Mg^{2+}$
C
$Cl^{-}$
D
$F^{-}$
Solution
(B) The species $Na^{+}$,$Mg^{2+}$,$F^{-}$,and $Cl^{-}$ are not all isoelectronic.
$Na^{+}$ $(10 \ e^{-})$,$Mg^{2+}$ $(10 \ e^{-})$,and $F^{-}$ $(10 \ e^{-})$ are isoelectronic,while $Cl^{-}$ $(18 \ e^{-})$ is larger due to an extra shell.
Among the isoelectronic species ($Na^{+}$,$Mg^{2+}$,$F^{-}$),the size decreases as the nuclear charge increases.
$Mg^{2+}$ has the highest atomic number $(Z=12)$ compared to $Na^{+}$ $(Z=11)$ and $F^{-}$ $(Z=9)$.
Therefore,$Mg^{2+}$ has the strongest attraction for electrons,resulting in the smallest ionic radius.
$Na^{+}$ $(10 \ e^{-})$,$Mg^{2+}$ $(10 \ e^{-})$,and $F^{-}$ $(10 \ e^{-})$ are isoelectronic,while $Cl^{-}$ $(18 \ e^{-})$ is larger due to an extra shell.
Among the isoelectronic species ($Na^{+}$,$Mg^{2+}$,$F^{-}$),the size decreases as the nuclear charge increases.
$Mg^{2+}$ has the highest atomic number $(Z=12)$ compared to $Na^{+}$ $(Z=11)$ and $F^{-}$ $(Z=9)$.
Therefore,$Mg^{2+}$ has the strongest attraction for electrons,resulting in the smallest ionic radius.
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18
ChemistryEasyMCQAIPMT · 1996
Which of the following is largest?
A
$Cl^{-}$
B
$S^{2-}$
C
$Na^{+}$
D
$F^{-}$
Solution
(B) All the given ions are isoelectronic,meaning they all have $18$ electrons ($Na^{+}$ has $10$ electrons,so let us compare the isoelectronic series $S^{2-}, Cl^{-}, K^{+}, Ca^{2+}$).
For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number) increases.
Atomic numbers are: $S$ $(16)$,$Cl$ $(17)$,$Na$ $(11)$,$F$ $(9)$.
$S^{2-}$ has the lowest nuclear charge $(Z = 16)$ among the given options $S^{2-}$ and $Cl^{-}$,therefore it has the largest ionic radius.
For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number) increases.
Atomic numbers are: $S$ $(16)$,$Cl$ $(17)$,$Na$ $(11)$,$F$ $(9)$.
$S^{2-}$ has the lowest nuclear charge $(Z = 16)$ among the given options $S^{2-}$ and $Cl^{-}$,therefore it has the largest ionic radius.
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19
ChemistryDifficultMCQAIPMT · 1996
Which one of the elements has the maximum electron affinity?
A
$F$
B
$Cl$
C
$Br$
D
$I$
Solution
(B) The correct answer is $(B)$.
Although $F$ has the highest electronegativity,its electron affinity is lower than that of $Cl$ due to its very small atomic size.
In $F$,the incoming electron experiences significant inter-electronic repulsions from the already present electrons in the small $2p$ subshell.
$Cl$,being larger,can accommodate the incoming electron more easily,resulting in a higher energy release (more negative electron gain enthalpy).
Although $F$ has the highest electronegativity,its electron affinity is lower than that of $Cl$ due to its very small atomic size.
In $F$,the incoming electron experiences significant inter-electronic repulsions from the already present electrons in the small $2p$ subshell.
$Cl$,being larger,can accommodate the incoming electron more easily,resulting in a higher energy release (more negative electron gain enthalpy).
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20
ChemistryEasyMCQAIPMT · 1996
The formula for calcium chlorite is
A
$Ca(ClO_4)_2$
B
$Ca(ClO_3)_2$
C
$CaCl(O_2)$
D
$Ca(ClO_2)_2$
Solution
(D) The chemical formula for calcium chlorite is derived from the calcium ion $(Ca^{2+})$ and the chlorite ion $(ClO_2^-)$.
To balance the charges,two chlorite ions are required for every one calcium ion,resulting in the formula $Ca(ClO_2)_2$.
Calcium chlorite is a white granular solid.
To balance the charges,two chlorite ions are required for every one calcium ion,resulting in the formula $Ca(ClO_2)_2$.
Calcium chlorite is a white granular solid.
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21
ChemistryMediumMCQAIPMT · 1996
Which of the following bonds has the highest bond energy?
A
$Se-Se$
B
$Te-Te$
C
$S-S$
D
$O-O$
Solution
(C) The bond energy depends on the size of the atoms and the extent of orbital overlap. $O-O$ bond has the lowest bond energy due to high inter-electronic repulsion between the lone pairs of small oxygen atoms. As we move down the group,the atomic size increases,which decreases the bond energy. However,the $S-S$ bond is stronger than $Se-Se$ and $Te-Te$ due to better orbital overlap compared to the larger atoms. Therefore,among the given options,the $S-S$ bond has the highest bond energy.
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22
ChemistryMediumMCQAIPMT · 1996
When thiosulphate ion is oxidised by iodine,which one of the following ions is produced?
A
$SO_3^{2-}$
B
$SO_4^{2-}$
C
$S_4O_6^{2-}$ (Tetrathionate)
D
$S_2O_6^{2-}$
Solution
(C) The reaction between thiosulphate ion $(S_2O_3^{2-})$ and iodine $(I_2)$ is a redox reaction.
In this reaction,the thiosulphate ion is oxidised to the tetrathionate ion $(S_4O_6^{2-})$,while iodine is reduced to iodide ion $(I^-)$.
The balanced chemical equation is:
$2S_2O_3^{2-} + I_2 \rightarrow S_4O_6^{2-} + 2I^-$
Therefore,the correct ion produced is the tetrathionate ion $(S_4O_6^{2-})$.
In this reaction,the thiosulphate ion is oxidised to the tetrathionate ion $(S_4O_6^{2-})$,while iodine is reduced to iodide ion $(I^-)$.
The balanced chemical equation is:
$2S_2O_3^{2-} + I_2 \rightarrow S_4O_6^{2-} + 2I^-$
Therefore,the correct ion produced is the tetrathionate ion $(S_4O_6^{2-})$.
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23
ChemistryMediumMCQAIPMT · 1996
The $IUPAC$ name of $(CH_3)_2CH-CH_2-CH_2Br$ is
A
$1-$bromopentane
B
$2-$methyl$-4-$bromobutane
C
$1-$bromo$-3-$methylbutane
D
$2-$methyl$-3-$bromopropane
Solution
(C) To determine the $IUPAC$ name,we identify the longest carbon chain containing the functional group.
The structure is $CH_3-CH(CH_3)-CH_2-CH_2Br$.
The longest chain has $4$ carbon atoms (butane).
Numbering starts from the carbon attached to the bromine atom to give it the lowest possible locant.
At position $1$,there is a bromo group,and at position $3$,there is a methyl group.
Therefore,the $IUPAC$ name is $1-$bromo$-3-$methylbutane.
The structure is $CH_3-CH(CH_3)-CH_2-CH_2Br$.
The longest chain has $4$ carbon atoms (butane).
Numbering starts from the carbon attached to the bromine atom to give it the lowest possible locant.
At position $1$,there is a bromo group,and at position $3$,there is a methyl group.
Therefore,the $IUPAC$ name is $1-$bromo$-3-$methylbutane.
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24
ChemistryMediumMCQAIPMT · 1996
The number of isomers of $C_4H_{10}$ is:
A
$2$
B
$3$
C
$4$
D
Isomerism does not exist
Solution
(A) The molecular formula $C_4H_{10}$ corresponds to an alkane.
There are two structural isomers for this formula:
$1$. $n$-Butane: $CH_3-CH_2-CH_2-CH_3$
$2$. Isobutane ($2$-methylpropane): $CH_3-CH(CH_3)-CH_3$
Therefore,the total number of isomers is $2$.
There are two structural isomers for this formula:
$1$. $n$-Butane: $CH_3-CH_2-CH_2-CH_3$
$2$. Isobutane ($2$-methylpropane): $CH_3-CH(CH_3)-CH_3$
Therefore,the total number of isomers is $2$.
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25
ChemistryMediumMCQAIPMT · 1996
The reaction $CH_2 = CH - CH_3 + HBr \rightarrow CH_3CHBr - CH_3$ is
A
Nucleophilic addition
B
Electrophilic addition
C
Electrophilic substitution
D
Free radical addition
Solution
(B) In this reaction,$HBr$ undergoes heterolytic fission to produce an electrophile $H^{+}$:
$HBr \rightarrow H^{+} + Br^{-}$
The electrophile $H^{+}$ attacks the double bond of propene to form a stable carbocation intermediate $(CH_3-\stackrel{+}{C}H - CH_3)$.
Subsequently,the nucleophile $Br^{-}$ attacks the carbocation to form $2$-bromopropane.
Since the initial step involves the attack of an electrophile on the alkene,this is an electrophilic addition reaction.
$HBr \rightarrow H^{+} + Br^{-}$
The electrophile $H^{+}$ attacks the double bond of propene to form a stable carbocation intermediate $(CH_3-\stackrel{+}{C}H - CH_3)$.
Subsequently,the nucleophile $Br^{-}$ attacks the carbocation to form $2$-bromopropane.
Since the initial step involves the attack of an electrophile on the alkene,this is an electrophilic addition reaction.
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26
ChemistryMediumMCQAIPMT · 1996
The bond length between an $sp^{3}$ hybridized carbon atom and another carbon atom is minimum in:
A
Propane
B
Butane
C
Propene
D
Propyne
Solution
(D) The bond length depends on the hybridization of the carbon atoms involved in the bond.
$sp^{3}-sp^{3}$ single bond length is $1.54 \, \mathring{A}$.
$sp^{2}-sp^{2}$ double bond length is $1.34 \, \mathring{A}$.
$sp-sp$ triple bond length is $1.20 \, \mathring{A}$.
In propyne $(CH_{3}-C \equiv CH)$,there is a $C \equiv C$ bond,which has the shortest bond length among the given options.
$sp^{3}-sp^{3}$ single bond length is $1.54 \, \mathring{A}$.
$sp^{2}-sp^{2}$ double bond length is $1.34 \, \mathring{A}$.
$sp-sp$ triple bond length is $1.20 \, \mathring{A}$.
In propyne $(CH_{3}-C \equiv CH)$,there is a $C \equiv C$ bond,which has the shortest bond length among the given options.
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27
ChemistryMediumMCQAIPMT · 1996
The electrophile in the case of chlorination of benzene in the presence of $FeCl_3$ is
A
$Cl^{+}$
B
$Cl^{-}$
C
$Cl$
D
$FeCl_3$
Solution
(A) In the chlorination of benzene,$FeCl_3$ acts as a Lewis acid catalyst.
It reacts with $Cl_2$ to generate the electrophile $Cl^{+}$ (chloronium ion).
The reaction is: $Cl_2 + FeCl_3 \rightarrow Cl^{+} + [FeCl_4]^{-}$.
Therefore,the correct option is $A$.
It reacts with $Cl_2$ to generate the electrophile $Cl^{+}$ (chloronium ion).
The reaction is: $Cl_2 + FeCl_3 \rightarrow Cl^{+} + [FeCl_4]^{-}$.
Therefore,the correct option is $A$.
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28
ChemistryDifficultMCQAIPMT · 1996
$KMnO_4$ reacts with oxalic acid according to the equation:
$2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \to 2Mn^{2+} + 10CO_2 + 8H_2O$
Here,$20 \ mL$ of $0.1 \ M$ $KMnO_4$ is equivalent to:
$2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \to 2Mn^{2+} + 10CO_2 + 8H_2O$
Here,$20 \ mL$ of $0.1 \ M$ $KMnO_4$ is equivalent to:
A
$20 \ mL$ of $0.5 \ M$ $C_2H_2O_4$
B
$50 \ mL$ of $0.1 \ M$ $C_2H_2O_4$
C
$50 \ mL$ of $0.5 \ M$ $C_2H_2O_4$
D
$20 \ mL$ of $0.1 \ M$ $C_2H_2O_4$
Solution
(B) From the balanced chemical equation,$2 \ \text{moles}$ of $MnO_4^-$ react with $5 \ \text{moles}$ of $C_2O_4^{2-}$.
The number of moles of $KMnO_4$ used is $n = M \times V = 0.1 \ \text{mol/L} \times 0.020 \ \text{L} = 0.002 \ \text{moles}$.
According to the stoichiometry,$2 \ \text{moles}$ of $KMnO_4$ require $5 \ \text{moles}$ of oxalic acid $(C_2H_2O_4)$.
Therefore,$0.002 \ \text{moles}$ of $KMnO_4$ require $(5/2) \times 0.002 = 0.005 \ \text{moles}$ of $C_2H_2O_4$.
Checking the options:
For option $B$: $n = 0.1 \ \text{M} \times 0.050 \ \text{L} = 0.005 \ \text{moles}$.
Thus,$50 \ \text{mL}$ of $0.1 \ \text{M}$ $C_2H_2O_4$ is equivalent.
The number of moles of $KMnO_4$ used is $n = M \times V = 0.1 \ \text{mol/L} \times 0.020 \ \text{L} = 0.002 \ \text{moles}$.
According to the stoichiometry,$2 \ \text{moles}$ of $KMnO_4$ require $5 \ \text{moles}$ of oxalic acid $(C_2H_2O_4)$.
Therefore,$0.002 \ \text{moles}$ of $KMnO_4$ require $(5/2) \times 0.002 = 0.005 \ \text{moles}$ of $C_2H_2O_4$.
Checking the options:
For option $B$: $n = 0.1 \ \text{M} \times 0.050 \ \text{L} = 0.005 \ \text{moles}$.
Thus,$50 \ \text{mL}$ of $0.1 \ \text{M}$ $C_2H_2O_4$ is equivalent.
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29
ChemistryMCQAIPMT · 1996
Which of the following is a micronutrient for plants?
A
Calcium
B
Magnesium
C
Manganese
D
Nitrogen
Solution
(C) Essential elements are classified into two categories based on their quantitative requirements: macronutrients and micronutrients.
$1$. Macronutrients are required in large amounts (usually $ > 10 \text{ mmol kg}^{-1}$ of dry matter). Examples include $C, H, O, N, P, K, S, Ca,$ and $Mg$.
$2$. Micronutrients or trace elements are required in very small amounts (usually $ < 10 \text{ mmol kg}^{-1}$ of dry matter). Examples include $Fe, Mn, Cu, Mo, Zn, B, Cl,$ and $Ni$.
Among the given options, $Ca, Mg,$ and $N$ are macronutrients, while $Mn$ (Manganese) is a micronutrient.
$1$. Macronutrients are required in large amounts (usually $ > 10 \text{ mmol kg}^{-1}$ of dry matter). Examples include $C, H, O, N, P, K, S, Ca,$ and $Mg$.
$2$. Micronutrients or trace elements are required in very small amounts (usually $ < 10 \text{ mmol kg}^{-1}$ of dry matter). Examples include $Fe, Mn, Cu, Mo, Zn, B, Cl,$ and $Ni$.
Among the given options, $Ca, Mg,$ and $N$ are macronutrients, while $Mn$ (Manganese) is a micronutrient.
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30
ChemistryMCQAIPMT · 1996
Which nitrogenous base is present in $RNA$ but not in $DNA$?
A
Uracil
B
Thymine
C
Guanine
D
Cytosine
Solution
(A) In nucleic acids,nitrogenous bases are categorized into purines and pyrimidines.
$DNA$ contains Adenine $(A)$,Guanine $(G)$,Cytosine $(C)$,and Thymine $(T)$.
$RNA$ contains Adenine $(A)$,Guanine $(G)$,Cytosine $(C)$,and Uracil $(U)$.
Therefore,Uracil is the nitrogenous base that is present in $RNA$ but is replaced by Thymine in $DNA$.
$DNA$ contains Adenine $(A)$,Guanine $(G)$,Cytosine $(C)$,and Thymine $(T)$.
$RNA$ contains Adenine $(A)$,Guanine $(G)$,Cytosine $(C)$,and Uracil $(U)$.
Therefore,Uracil is the nitrogenous base that is present in $RNA$ but is replaced by Thymine in $DNA$.
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31
ChemistryMCQAIPMT · 1996
Which of the following is not a Lewis acid?
A
$SiF_4$
B
$FeCl_3$
C
$BF_3$
D
$C_2H_4$
Solution
(D) Lewis acid is a chemical species that can accept an electron pair.
$SiF_4$ has an empty $d$-orbital on the silicon atom,allowing it to accept electron pairs.
$FeCl_3$ has an incomplete octet on the iron atom,making it an electron-deficient species and a Lewis acid.
$BF_3$ has an incomplete octet on the boron atom (only $6$ electrons),making it a classic Lewis acid.
$C_2H_4$ (ethene) has all its atoms with complete octets and no vacant orbitals to accept an electron pair,thus it is not a Lewis acid.
$SiF_4$ has an empty $d$-orbital on the silicon atom,allowing it to accept electron pairs.
$FeCl_3$ has an incomplete octet on the iron atom,making it an electron-deficient species and a Lewis acid.
$BF_3$ has an incomplete octet on the boron atom (only $6$ electrons),making it a classic Lewis acid.
$C_2H_4$ (ethene) has all its atoms with complete octets and no vacant orbitals to accept an electron pair,thus it is not a Lewis acid.
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32
ChemistryMCQAIPMT · 1996
During the first meiotic division,the exchange of genetic material between the chromatids of homologous chromosomes occurs. What is this process called?
A
Transformation
B
Chiasmata
C
Crossing over
D
Synapsis
Solution
(C) During the $Pachytene$ stage of $Prophase-I$ of meiosis,the phenomenon of crossing over takes place.
Crossing over is the exchange of genetic material between non-sister chromatids of homologous chromosomes.
This process leads to genetic recombination,which is essential for variation in sexually reproducing organisms.
$Synapsis$ refers to the pairing of homologous chromosomes,while $Chiasmata$ are the $X$-shaped structures formed at the sites where crossing over has occurred.
Crossing over is the exchange of genetic material between non-sister chromatids of homologous chromosomes.
This process leads to genetic recombination,which is essential for variation in sexually reproducing organisms.
$Synapsis$ refers to the pairing of homologous chromosomes,while $Chiasmata$ are the $X$-shaped structures formed at the sites where crossing over has occurred.
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33
ChemistryMCQAIPMT · 1996
Which of the following bonds has the highest energy?
A
$Se-Se$
B
$Te-Te$
C
$S-S$
D
$O-O$
Solution
(C) The bond energy depends on the size of the atoms involved in the bond. Smaller atoms form stronger bonds due to better orbital overlap. Among the given elements $(O, S, Se, Te)$,oxygen $(O)$ is the smallest. However,the $O-O$ single bond is exceptionally weak due to the high inter-electronic repulsion between the lone pairs of the small oxygen atoms. Therefore,the bond energy trend for single bonds in the group is $S-S > Se-Se > Te-Te > O-O$. Thus,the $S-S$ bond has the highest bond energy among the given options.
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34
ChemistryMCQAIPMT · 1996
Which of the following is not a Lewis acid?
A
$SiF_4$
B
$FeCl_3$
C
$BF_3$
D
$C_2H_4$
Solution
(D) Lewis acid is defined as an electron pair acceptor.
$SiF_4$ has an empty $d$-orbital on the $Si$ atom,allowing it to accept electron pairs.
$FeCl_3$ has an incomplete octet on the $Fe$ atom,making it an electron pair acceptor.
$BF_3$ has an incomplete octet on the $B$ atom (only $6$ electrons),making it a classic Lewis acid.
$C_2H_4$ (ethene) has a $\pi$-bond and acts as a Lewis base because it can donate electron density from its double bond,not as a Lewis acid.
$SiF_4$ has an empty $d$-orbital on the $Si$ atom,allowing it to accept electron pairs.
$FeCl_3$ has an incomplete octet on the $Fe$ atom,making it an electron pair acceptor.
$BF_3$ has an incomplete octet on the $B$ atom (only $6$ electrons),making it a classic Lewis acid.
$C_2H_4$ (ethene) has a $\pi$-bond and acts as a Lewis base because it can donate electron density from its double bond,not as a Lewis acid.
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35
ChemistryMCQAIPMT · 1996
$A$ nucleus ruptures into two nuclear parts which have their velocity ratio equal to $2 : 1$. What will be the ratio of their nuclear size (nuclear radius)?
A
$2^{1/3} : 1$
B
$1 : 2^{1/3}$
C
$3^{1/2} : 1$
D
$1 : 3^{1/2}$
Solution
(B) According to the law of conservation of linear momentum,$m_1v_1 = m_2v_2$.
Given the velocity ratio $v_1 : v_2 = 2 : 1$,we have $\frac{v_1}{v_2} = 2$.
From the momentum conservation equation,the mass ratio is $\frac{m_1}{m_2} = \frac{v_2}{v_1} = \frac{1}{2}$.
Since the mass number $A$ is proportional to the mass of the nucleus,$\frac{A_1}{A_2} = \frac{1}{2}$.
The nuclear radius is given by $R = R_0 A^{1/3}$.
Therefore,the ratio of their nuclear radii is $\frac{R_1}{R_2} = \left(\frac{A_1}{A_2}\right)^{1/3} = \left(\frac{1}{2}\right)^{1/3} = 1 : 2^{1/3}$.
Given the velocity ratio $v_1 : v_2 = 2 : 1$,we have $\frac{v_1}{v_2} = 2$.
From the momentum conservation equation,the mass ratio is $\frac{m_1}{m_2} = \frac{v_2}{v_1} = \frac{1}{2}$.
Since the mass number $A$ is proportional to the mass of the nucleus,$\frac{A_1}{A_2} = \frac{1}{2}$.
The nuclear radius is given by $R = R_0 A^{1/3}$.
Therefore,the ratio of their nuclear radii is $\frac{R_1}{R_2} = \left(\frac{A_1}{A_2}\right)^{1/3} = \left(\frac{1}{2}\right)^{1/3} = 1 : 2^{1/3}$.
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36
ChemistryMCQAIPMT · 1996
$A$ nucleus ruptures into two nuclear parts which have their velocity ratio equal to $2:1$. What will be the ratio of their nuclear size (nuclear radius)?
A
$2^{1/3} : 1$
B
$1 : 2^{1/3}$
C
$3^{1/2} : 1$
D
$1 : 3^{1/2}$
Solution
(B) According to the law of conservation of linear momentum,the total momentum before rupture is zero,so the magnitudes of the momenta of the two parts must be equal:
$m_1 v_1 = m_2 v_2$
$\frac{m_1}{m_2} = \frac{v_2}{v_1}$
Given the velocity ratio $\frac{v_1}{v_2} = \frac{2}{1}$,we have $\frac{v_2}{v_1} = \frac{1}{2}$.
Since the mass of a nucleus is proportional to its mass number $A$ $(m \propto A)$,we have $\frac{A_1}{A_2} = \frac{m_1}{m_2} = \frac{1}{2}$.
The nuclear radius $R$ is related to the mass number $A$ by the formula $R = R_0 A^{1/3}$.
Therefore,the ratio of the radii is $\frac{R_1}{R_2} = \left(\frac{A_1}{A_2}\right)^{1/3} = \left(\frac{1}{2}\right)^{1/3} = 1 : 2^{1/3}$.
$m_1 v_1 = m_2 v_2$
$\frac{m_1}{m_2} = \frac{v_2}{v_1}$
Given the velocity ratio $\frac{v_1}{v_2} = \frac{2}{1}$,we have $\frac{v_2}{v_1} = \frac{1}{2}$.
Since the mass of a nucleus is proportional to its mass number $A$ $(m \propto A)$,we have $\frac{A_1}{A_2} = \frac{m_1}{m_2} = \frac{1}{2}$.
The nuclear radius $R$ is related to the mass number $A$ by the formula $R = R_0 A^{1/3}$.
Therefore,the ratio of the radii is $\frac{R_1}{R_2} = \left(\frac{A_1}{A_2}\right)^{1/3} = \left(\frac{1}{2}\right)^{1/3} = 1 : 2^{1/3}$.
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37
ChemistryMCQAIPMT · 1996
$A$ beam of electrons is moving with constant velocity in a region having electric and magnetic fields of strength $20 \; Vm^{-1}$ and $0.5 \; T$ at right angles to the direction of motion of the electrons. What is the velocity (in $m/s$) of the electrons?
A
$5.5$
B
$40$
C
$8$
D
$20$
Solution
(B) Given: Electric field $E = 20 \; V/m$ and magnetic field $B = 0.5 \; T$.
The force exerted on an electron by an electric field is $F_e = eE$.
The force exerted on an electron by a magnetic field is $F_m = evB$ (since the velocity is perpendicular to the magnetic field).
Since the electron moves with a constant velocity,the net force acting on it must be zero. Therefore,the electric force must balance the magnetic force:
$eE = evB$
Solving for velocity $v$:
$v = \frac{E}{B}$
Substituting the given values:
$v = \frac{20}{0.5} = 40 \; m/s$.
The force exerted on an electron by an electric field is $F_e = eE$.
The force exerted on an electron by a magnetic field is $F_m = evB$ (since the velocity is perpendicular to the magnetic field).
Since the electron moves with a constant velocity,the net force acting on it must be zero. Therefore,the electric force must balance the magnetic force:
$eE = evB$
Solving for velocity $v$:
$v = \frac{E}{B}$
Substituting the given values:
$v = \frac{20}{0.5} = 40 \; m/s$.
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38
ChemistryMCQAIPMT · 1996
Which of the following is not a Lewis acid?
A
$SiF_4$
B
$FeCl_3$
C
$BF_3$
D
$C_2H_4$
Solution
(D) Lewis acid is a substance that can accept a pair of electrons.
$SiF_4$ has an empty $d$-orbital on the $Si$ atom,allowing it to accept electrons.
$FeCl_3$ has an incomplete octet on the $Fe$ atom,making it an electron-pair acceptor.
$BF_3$ has an incomplete octet on the $B$ atom (only $6$ electrons),making it a classic Lewis acid.
$C_2H_4$ (ethene) has a double bond and all atoms have complete octets; it acts as a Lewis base due to the electron density in the $\pi$-bond,not as a Lewis acid.
$SiF_4$ has an empty $d$-orbital on the $Si$ atom,allowing it to accept electrons.
$FeCl_3$ has an incomplete octet on the $Fe$ atom,making it an electron-pair acceptor.
$BF_3$ has an incomplete octet on the $B$ atom (only $6$ electrons),making it a classic Lewis acid.
$C_2H_4$ (ethene) has a double bond and all atoms have complete octets; it acts as a Lewis base due to the electron density in the $\pi$-bond,not as a Lewis acid.
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39
ChemistryMCQAIPMT · 1996
If we remove half of the forest cover of the Earth,the crisis that will occur is:
A
Many species would become extinct
B
Population,pollution,and ecological imbalance will rise
C
Energy crisis will commence
D
The remaining forest will correct the imbalance
Solution
(A) Forests are vital to the Earth's ecosystem,covering a significant portion of the land surface. They play a crucial role in maintaining ecological balance by regulating the climate,providing habitats for countless species,and sequestering carbon dioxide. If half of the forest cover were removed,it would lead to massive habitat loss,causing many species to become extinct. Furthermore,it would trigger severe ecological imbalances,increase pollution levels,and disrupt global climate patterns,ultimately leading to a catastrophic environmental crisis.
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40
ChemistryEasyMCQAIPMT · 1996
Which of the following oxides is least acidic?
A
$P_4O_6$
B
$P_4O_{10}$
C
$As_4O_6$
D
$As_4O_{10}$
Solution
(C) The acidic character of oxides depends on the oxidation state of the central atom and its position in the periodic table.
$1$. Acidic character increases with an increase in the oxidation state of the central atom.
$2$. Acidic character decreases down the group as the metallic character increases.
Comparing the oxidation states: $P$ in $P_4O_6$ is $+3$,$P$ in $P_4O_{10}$ is $+5$,$As$ in $As_4O_6$ is $+3$,and $As$ in $As_4O_{10}$ is $+5$.
Since $As$ is less electronegative than $P$,the oxides of $As$ are less acidic than the corresponding oxides of $P$.
Among the given options,$As_4O_6$ has the lowest oxidation state $(+3)$ and belongs to a lower period than $P$,making it the least acidic.
$1$. Acidic character increases with an increase in the oxidation state of the central atom.
$2$. Acidic character decreases down the group as the metallic character increases.
Comparing the oxidation states: $P$ in $P_4O_6$ is $+3$,$P$ in $P_4O_{10}$ is $+5$,$As$ in $As_4O_6$ is $+3$,and $As$ in $As_4O_{10}$ is $+5$.
Since $As$ is less electronegative than $P$,the oxides of $As$ are less acidic than the corresponding oxides of $P$.
Among the given options,$As_4O_6$ has the lowest oxidation state $(+3)$ and belongs to a lower period than $P$,making it the least acidic.
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41
ChemistryMediumMCQAIPMT · 1996
The basic character of hydrides of the $V$-group elements decreases in the order:
A
$SbH_3 > PH_3 > AsH_3 > NH_3$
B
$NH_3 > SbH_3 > PH_3 > AsH_3$
C
$NH_3 > PH_3 > AsH_3 > SbH_3$
D
$SbH_3 > AsH_3 > PH_3 > NH_3$
Solution
(C) The correct order is $NH_3 > PH_3 > AsH_3 > SbH_3$.
As we move down the group,the atomic size of the central atom increases.
This leads to the dispersion of the lone pair over a larger volume,which decreases the electron density on the central atom.
Consequently,the availability of the lone pair for donation decreases,and the basic character decreases.
As we move down the group,the atomic size of the central atom increases.
This leads to the dispersion of the lone pair over a larger volume,which decreases the electron density on the central atom.
Consequently,the availability of the lone pair for donation decreases,and the basic character decreases.
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42
ChemistryMediumMCQAIPMT · 1996
Which of the following substances is used as an antiknock compound?
A
Lead tetrachloride
B
Lead acetate
C
Zinc ethyl
D
Tetraethyl lead $(TEL)$
Solution
(D) The correct answer is $(d)$.
Tetraethyl lead $(TEL)$ is an antiknock compound.
When mixed with petrol,it improves the octane number of the fuel.
This process decreases the knocking in the cylinder of an internal combustion engine.
Tetraethyl lead $(TEL)$ is an antiknock compound.
When mixed with petrol,it improves the octane number of the fuel.
This process decreases the knocking in the cylinder of an internal combustion engine.
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43
ChemistryMediumMCQAIPMT · 1996
The relationship between osmotic pressure at $273 \ K$ when $10 \ g$ glucose $(P_1)$,$10 \ g$ urea $(P_2)$,and $10 \ g$ sucrose $(P_3)$ are dissolved in $250 \ mL$ of water is:
A
$P_1 > P_2 > P_3$
B
$P_3 > P_1 > P_2$
C
$P_2 > P_1 > P_3$
D
$P_2 > P_3 > P_1$
Solution
(C) The formula for osmotic pressure is $P = \frac{w}{MV} RT$.
Since $w$ (mass),$V$ (volume),and $T$ (temperature) are constant for all three solutions,the osmotic pressure is inversely proportional to the molar mass $(M)$ of the solute,i.e.,$P \propto \frac{1}{M}$.
The molar masses are:
Urea $(NH_2CONH_2) = 60 \ g/mol$
Glucose $(C_6H_{12}O_6) = 180 \ g/mol$
Sucrose $(C_{12}H_{22}O_{11}) = 342 \ g/mol$
Since $60 < 180 < 342$,the order of osmotic pressure is $P_2 > P_1 > P_3$.
Since $w$ (mass),$V$ (volume),and $T$ (temperature) are constant for all three solutions,the osmotic pressure is inversely proportional to the molar mass $(M)$ of the solute,i.e.,$P \propto \frac{1}{M}$.
The molar masses are:
Urea $(NH_2CONH_2) = 60 \ g/mol$
Glucose $(C_6H_{12}O_6) = 180 \ g/mol$
Sucrose $(C_{12}H_{22}O_{11}) = 342 \ g/mol$
Since $60 < 180 < 342$,the order of osmotic pressure is $P_2 > P_1 > P_3$.
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44
ChemistryEasyMCQAIPMT · 1996
The number of atoms contained in one face-centred cubic $(FCC)$ unit cell of a monoatomic substance is:
A
$1$
B
$2$
C
$4$
D
$6$
Solution
(C) In a face-centred cubic $(FCC)$ unit cell,atoms are present at the corners and at the centre of each face.
Number of atoms at corners = $8 \times \frac{1}{8} = 1$.
Number of atoms at face centres = $6 \times \frac{1}{2} = 3$.
Total number of atoms = $1 + 3 = 4$.
Number of atoms at corners = $8 \times \frac{1}{8} = 1$.
Number of atoms at face centres = $6 \times \frac{1}{2} = 3$.
Total number of atoms = $1 + 3 = 4$.
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45
ChemistryMediumMCQAIPMT · 1996
An element (atomic mass $100 \ g/mol$) having $bcc$ structure has unit cell edge $400 \ pm$. Then density of the element is (in $g/cm^3$)
A
$10.376$
B
$5.188$
C
$7.289$
D
$2.144$
Solution
(B) For a $bcc$ structure, the number of atoms per unit cell $(n)$ is $2$.
Given: Atomic mass $(M)$ = $100 \ g/mol$, edge length $(a)$ = $400 \ pm = 400 \times 10^{-10} \ cm$, Avogadro's number $(N_A)$ = $6.022 \times 10^{23} \ mol^{-1}$.
The formula for density $(\rho)$ is $\rho = \frac{n \times M}{a^3 \times N_A}$.
Substituting the values: $\rho = \frac{2 \times 100}{(400 \times 10^{-10})^3 \times 6.022 \times 10^{23}}$.
$\rho = \frac{200}{64 \times 10^{-24} \times 6.022 \times 10^{23}} = \frac{200}{38.54} \approx 5.189 \ g/cm^3$.
Thus, the correct option is $(B)$.
Given: Atomic mass $(M)$ = $100 \ g/mol$, edge length $(a)$ = $400 \ pm = 400 \times 10^{-10} \ cm$, Avogadro's number $(N_A)$ = $6.022 \times 10^{23} \ mol^{-1}$.
The formula for density $(\rho)$ is $\rho = \frac{n \times M}{a^3 \times N_A}$.
Substituting the values: $\rho = \frac{2 \times 100}{(400 \times 10^{-10})^3 \times 6.022 \times 10^{23}}$.
$\rho = \frac{200}{64 \times 10^{-24} \times 6.022 \times 10^{23}} = \frac{200}{38.54} \approx 5.189 \ g/cm^3$.
Thus, the correct option is $(B)$.
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46
ChemistryMediumMCQAIPMT · 1996
The reaction $2FeCl_3 + SnCl_2 \to 2FeCl_2 + SnCl_4$ is an example of
A
First order reaction
B
Second order reaction
C
Third order reaction
D
None of these
Solution
(C) The rate law for the given reaction is determined experimentally as $r = k[FeCl_3]^2 [SnCl_2]^1$.
The order of the reaction is the sum of the powers of the concentration terms in the rate law expression.
Order $= 2 + 1 = 3$.
Therefore,it is a third-order reaction.
The order of the reaction is the sum of the powers of the concentration terms in the rate law expression.
Order $= 2 + 1 = 3$.
Therefore,it is a third-order reaction.
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47
ChemistryEasyMCQAIPMT · 1996
$5 \ A$ current is passed through a solution of zinc sulphate for $40 \ \text{minutes}$. Find the amount of zinc deposited at the cathode in $gm$.
A
$40.65$
B
$4.065$
C
$0.4065$
D
$65.04$
Solution
(B) According to Faraday's law of electrolysis,the mass deposited $W$ is given by $W = \frac{Z \times I \times t}{96500}$.
Here,$I = 5 \ A$,$t = 40 \times 60 \ s = 2400 \ s$,and the equivalent mass of zinc $(Zn^{2+})$ is $Z = \frac{65.38}{2} = 32.69 \ g/mol$.
Substituting the values: $W = \frac{32.69 \times 5 \times 2400}{96500}$.
$W = \frac{392280}{96500} \approx 4.065 \ g$.
Here,$I = 5 \ A$,$t = 40 \times 60 \ s = 2400 \ s$,and the equivalent mass of zinc $(Zn^{2+})$ is $Z = \frac{65.38}{2} = 32.69 \ g/mol$.
Substituting the values: $W = \frac{32.69 \times 5 \times 2400}{96500}$.
$W = \frac{392280}{96500} \approx 4.065 \ g$.
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48
ChemistryMediumMCQAIPMT · 1996
The standard oxidation potentials for the half-reactions are given as $Zn \to Zn^{2+} + 2e^{-}; E^o = +0.76 \ V$ and $Fe \to Fe^{2+} + 2e^{-}; E^o = +0.41 \ V$. The $EMF$ for the cell reaction $Fe^{2+} + Zn \to Zn^{2+} + Fe$ is ............ $V$.
A
$-0.35$
B
$+0.35$
C
$+1.17$
D
$-1.17$
Solution
(B) The given reactions are oxidation half-reactions. The standard reduction potentials $(E^o_{red})$ are the negative of the standard oxidation potentials $(E^o_{ox})$.
$E^o_{red}(Zn^{2+}/Zn) = -0.76 \ V$
$E^o_{red}(Fe^{2+}/Fe) = -0.41 \ V$
In the cell reaction $Fe^{2+} + Zn \to Zn^{2+} + Fe$,$Zn$ is oxidized (anode) and $Fe^{2+}$ is reduced (cathode).
$E^o_{cell} = E^o_{cathode} - E^o_{anode}$
$E^o_{cell} = E^o_{red}(Fe^{2+}/Fe) - E^o_{red}(Zn^{2+}/Zn)$
$E^o_{cell} = -0.41 \ V - (-0.76 \ V)$
$E^o_{cell} = -0.41 \ V + 0.76 \ V = +0.35 \ V$.
$E^o_{red}(Zn^{2+}/Zn) = -0.76 \ V$
$E^o_{red}(Fe^{2+}/Fe) = -0.41 \ V$
In the cell reaction $Fe^{2+} + Zn \to Zn^{2+} + Fe$,$Zn$ is oxidized (anode) and $Fe^{2+}$ is reduced (cathode).
$E^o_{cell} = E^o_{cathode} - E^o_{anode}$
$E^o_{cell} = E^o_{red}(Fe^{2+}/Fe) - E^o_{red}(Zn^{2+}/Zn)$
$E^o_{cell} = -0.41 \ V - (-0.76 \ V)$
$E^o_{cell} = -0.41 \ V + 0.76 \ V = +0.35 \ V$.
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49
ChemistryDifficultMCQAIPMT · 1996
The size of colloidal particles varies from:
A
$10^{-9} \ m$ to $10^{-7} \ m$
B
$10^{-9} \ m$ to $10^{-17} \ m$
C
$10^{-5} \ m$ to $10^{-7} \ m$
D
$10^{-4} \ m$ to $10^{-10} \ m$
Solution
(A) The size of colloidal particles is intermediate between true solutions and suspensions.
The range of particle size for colloids is typically $1 \ nm$ to $1000 \ nm$.
Converting this to meters: $1 \ nm = 10^{-9} \ m$ and $1000 \ nm = 10^{-6} \ m$.
Among the given options,the range $10^{-9} \ m$ to $10^{-7} \ m$ (which is $1 \ nm$ to $100 \ nm$) represents the standard colloidal range.
Therefore,the correct option is $A$.
The range of particle size for colloids is typically $1 \ nm$ to $1000 \ nm$.
Converting this to meters: $1 \ nm = 10^{-9} \ m$ and $1000 \ nm = 10^{-6} \ m$.
Among the given options,the range $10^{-9} \ m$ to $10^{-7} \ m$ (which is $1 \ nm$ to $100 \ nm$) represents the standard colloidal range.
Therefore,the correct option is $A$.
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50
ChemistryEasyMCQAIPMT · 1996
The general electronic configuration of transition elements is
A
$(n - 1)d^{1 - 5}$
B
$(n - 1)d^{1 - 10}ns^1$
C
$(n - 1)d^{1 - 10}ns^{1 - 2}$
D
$ns^2(n - 1)d^{10}$
Solution
(C) Generally,$d$-block elements are called transition elements because they contain an inner partially filled $d$-subshell.
Thus,their general electronic configuration is represented as $(n - 1)d^{1 - 10}ns^{1 - 2}$.
Thus,their general electronic configuration is represented as $(n - 1)d^{1 - 10}ns^{1 - 2}$.
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51
ChemistryMediumMCQAIPMT · 1996
$A$ transition element $X$ has a configuration $[Ar]3d^4$ in its $+3$ oxidation state. Its atomic number is
A
$25$
B
$26$
C
$22$
D
$19$
Solution
(A) The electronic configuration of the ion $X^{+3}$ is $[Ar]3d^4$.
The total number of electrons in $X^{+3}$ is $18 + 4 = 22$.
Since the element $X$ is in the $+3$ oxidation state,it has lost $3$ electrons to form the $X^{+3}$ ion.
Therefore,the atomic number of the neutral element $X$ is $22 + 3 = 25$.
The total number of electrons in $X^{+3}$ is $18 + 4 = 22$.
Since the element $X$ is in the $+3$ oxidation state,it has lost $3$ electrons to form the $X^{+3}$ ion.
Therefore,the atomic number of the neutral element $X$ is $22 + 3 = 25$.
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52
ChemistryDifficultMCQAIPMT · 1996
When calomel reacts with $NH_4OH$,we get
A
$HgNH_2Cl$
B
$NH_2-Hg-Hg-Cl$
C
$Hg_2O$
D
$HgO$
Solution
(A) The reaction of calomel $(Hg_2Cl_2)$ with ammonium hydroxide $(NH_4OH)$ is a disproportionation reaction.
$Hg_2Cl_2 + 2NH_4OH \to Hg + Hg(NH_2)Cl + NH_4Cl + 2H_2O$.
The product $Hg(NH_2)Cl$ is known as amino mercuric chloride,which is a black precipitate.
Therefore,the correct option is $A$.
$Hg_2Cl_2 + 2NH_4OH \to Hg + Hg(NH_2)Cl + NH_4Cl + 2H_2O$.
The product $Hg(NH_2)Cl$ is known as amino mercuric chloride,which is a black precipitate.
Therefore,the correct option is $A$.
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53
ChemistryMediumMCQAIPMT · 1996
The oxidation of toluene to benzaldehyde by chromyl chloride $(CrO_2Cl_2)$ is called:
A
Cannizzaro reaction
B
Wurtz reaction
C
Etard reaction
D
Reimer-Tiemann reaction
Solution
(C) The reaction of toluene with chromyl chloride $(CrO_2Cl_2)$ in a suitable solvent like $CS_2$ or $CCl_4$ followed by hydrolysis yields benzaldehyde. This specific oxidation reaction is known as the Etard reaction.
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54
ChemistryMediumMCQAIPMT · 1996
Aldol condensation will not take place in
A
$HCHO$
B
$CH_3CH_2CHO$
C
$CH_3CHO$
D
$CH_3COCH_3$
Solution
(A) Aldol condensation requires the presence of at least one $\alpha$-hydrogen atom in the aldehyde or ketone.
$HCHO$ (Formaldehyde) does not contain any $\alpha$-carbon atom,and therefore,it lacks $\alpha$-hydrogen atoms.
Consequently,$HCHO$ cannot undergo aldol condensation.
$HCHO$ (Formaldehyde) does not contain any $\alpha$-carbon atom,and therefore,it lacks $\alpha$-hydrogen atoms.
Consequently,$HCHO$ cannot undergo aldol condensation.
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55
ChemistryMediumMCQAIPMT · 1996
Which of the following gives benzoic acid on oxidation?
A
Chlorophenol
B
Chlorotoluene
C
Chlorobenzene
D
Benzyl chloride
Solution
(D) Benzyl chloride $(C_6H_5CH_2Cl)$ on oxidation with strong oxidizing agents like acidic $K_2Cr_2O_7$ undergoes oxidation to form benzoic acid $(C_6H_5COOH)$.
The reaction is as follows:
$C_6H_5CH_2Cl + [O] \xrightarrow{K_2Cr_2O_7 / H_2SO_4} C_6H_5COOH + HCl$
Therefore,the correct option is $(D)$.
The reaction is as follows:
$C_6H_5CH_2Cl + [O] \xrightarrow{K_2Cr_2O_7 / H_2SO_4} C_6H_5COOH + HCl$
Therefore,the correct option is $(D)$.
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