Question 14 Marks
Answer
View full question & answer→(B)
Sol.
Sol.
Questions
SECTION - A [CHEMISTY - MCQ]
Take a timed test
20 questions · 17 auto-graded MCQ + 3 self-marked written.
MCQ 24 Marks
- ✓Both statement I and statement II are true
- BBoth statement I and statement II are false
- CStatement I is false but statement II is true
- DStatement I is true but statement II is false
Answer
View full question & answer→Correct option: A.
Both statement I and statement II are true
(A)
Sol.
Sol.
MCQ 34 Marks
Which of the following statement is true with respect to $\mathrm{H}_{2} \mathrm{O}, \mathrm{NH}_{3}$ and $\mathrm{CH}_{4}$ ?
A. The central atoms of all the molecules are $\mathrm{sp}^{3}$ hybridized.
B. The $\mathrm{H}-\mathrm{O}-\mathrm{H}, \mathrm{H}-\mathrm{N}-\mathrm{H}$ and $\mathrm{H}-\mathrm{C}-\mathrm{H}$ angles in the above molecules are $104.5^{\circ}, 107.5^{\circ}$ and $109.5^{\circ}$ respectively.
C. The increasing order of dipole moment is $\mathrm{CH}_{4}<\mathrm{NH}_{3}<\mathrm{H}_{2} \mathrm{O}$.
D. Both $\mathrm{H}_{2} \mathrm{O}$ and $\mathrm{NH}_{3}$ are Lewis acids and $\mathrm{CH}_{4}$ is a Lewis base
E. A solution of $\mathrm{NH}_{3}$ in $\mathrm{H}_{2} \mathrm{O}$ is basic. In this solution $\mathrm{NH}_{3}$ and $\mathrm{H}_{2} \mathrm{O}$ act as Lowry-Bronsted acid and base respectively.
Choose the correct answer from the options given below :
A. The central atoms of all the molecules are $\mathrm{sp}^{3}$ hybridized.
B. The $\mathrm{H}-\mathrm{O}-\mathrm{H}, \mathrm{H}-\mathrm{N}-\mathrm{H}$ and $\mathrm{H}-\mathrm{C}-\mathrm{H}$ angles in the above molecules are $104.5^{\circ}, 107.5^{\circ}$ and $109.5^{\circ}$ respectively.
C. The increasing order of dipole moment is $\mathrm{CH}_{4}<\mathrm{NH}_{3}<\mathrm{H}_{2} \mathrm{O}$.
D. Both $\mathrm{H}_{2} \mathrm{O}$ and $\mathrm{NH}_{3}$ are Lewis acids and $\mathrm{CH}_{4}$ is a Lewis base
E. A solution of $\mathrm{NH}_{3}$ in $\mathrm{H}_{2} \mathrm{O}$ is basic. In this solution $\mathrm{NH}_{3}$ and $\mathrm{H}_{2} \mathrm{O}$ act as Lowry-Bronsted acid and base respectively.
Choose the correct answer from the options given below :
- ✓A, B and C only
- BC, D and E only
- CA, D and E only
- DA, B, C and E only
Answer
View full question & answer→Correct option: A.
A, B and C only
(A)
Sol.
Sol.
MCQ 44 Marks
Consider the given plots of vapour pressure (VP) vs temperature (T/K) Which amongst the following options is correct graphical representation showing $\Delta \mathrm{T}_{\mathrm{f}}$, depression in the freezing point of solvent in a solution?
- A
- B
- ✓
- D
Answer
View full question & answer→Correct option: C.
(C)
Sol.
On adding non-volatile solute in a solvent, the freezing point of solution decreases.
$\mathrm{T}_{\mathrm{f}}<\mathrm{T}_{\mathrm{f}}^{0}$
F.P. of solution < F.P. of pure solvent Also V.P. of solution decreases on adding nonvolatile solute in a solvent.
Sol.
On adding non-volatile solute in a solvent, the freezing point of solution decreases.
$\mathrm{T}_{\mathrm{f}}<\mathrm{T}_{\mathrm{f}}^{0}$
F.P. of solution < F.P. of pure solvent Also V.P. of solution decreases on adding nonvolatile solute in a solvent.
Question 54 Marks
Answer
View full question & answer→(A)
Sol.
Sol.
MCQ 64 Marks
Which of the following arrangements with respect to their reactivity in nucleophilic addition reaction is correct?
- Abenzaldehyde $<$ acetophenone $<$ p-nitrobenzaldehyde $<$ p-tolualdehyde
- Bacetophenone $<$ benzaldehyde $<$ p-tolualdehyde $<$ p-nitrobenzaldehyde
- ✓acetophenone $<$ p-tolualdehyde < benzaldehyde < p-nitrobenzaldehyde
- Dp-nitrobenzaldehyde $<$ benzaldehyde $<$ p-tolualdehyde $<$ acetophenone
Answer
View full question & answer→Correct option: C.
acetophenone $<$ p-tolualdehyde < benzaldehyde < p-nitrobenzaldehyde
(C)
Sol.
Sol.
MCQ 74 Marks
For a reaction, $\mathrm{N}_{2} \mathrm{O}_{5(\mathrm{~g})} \rightarrow 2 \mathrm{NO}_{2(\mathrm{~g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})}$ in a constant volume container, no products were present initially. The final pressure of the system when $50 \%$ of reaction gets completed is
- A$7 / 2$ times of initial pressure
- B5 times of initial pressure
- C$5 / 2$ times of initial pressure
- ✓$7 / 4$ times of initial pressure
Answer
View full question & answer→Correct option: D.
$7 / 4$ times of initial pressure
(D)
Sol
Sol
MCQ 84 Marks
The large difference between the melting and boiling points of oxygen and sulphur may be explained on the basis of
- AAtomic size
- ✓Atomicity
- CElectronegativity
- DElectron gain enthalpy
Answer
View full question & answer→Correct option: B.
Atomicity
(B)
Sol. Oxygen exists as $\mathrm{O}_{2}($ Atomicity $=2)$
Sulphur exists as $\mathrm{S}_{8}($ Atomicity $=8)$
Hence, Melting point & Boiling point of sulphur are significantly large compared to oxygen.
Sol. Oxygen exists as $\mathrm{O}_{2}($ Atomicity $=2)$
Sulphur exists as $\mathrm{S}_{8}($ Atomicity $=8)$
Hence, Melting point & Boiling point of sulphur are significantly large compared to oxygen.
MCQ 94 Marks
Preparation of potassium permanganate from $\mathrm{MnO}_{2}$ involves two step process in which the $1^{\text {st }}$ step is a reaction with KOH and $\mathrm{KNO}_{3}$ to produce
- A$\mathrm{K}_{4}\left[\mathrm{Mn}(\mathrm{OH})_{6}\right]$
- B$\mathrm{K}_{3} \mathrm{MnO}_{4}$
- C$\mathrm{KMnO}_{4}$
- ✓$\mathrm{K}_{2} \mathrm{MnO}_{4}$
Answer
View full question & answer→Correct option: D.
$\mathrm{K}_{2} \mathrm{MnO}_{4}$
(D)
Sol. $\mathrm{MnO}_{2} \xrightarrow[\mathrm{KNO}_{3}, \Delta]{\mathrm{KOH}} \mathrm{K}_{2} \mathrm{MnO}_{4}$
Sol. $\mathrm{MnO}_{2} \xrightarrow[\mathrm{KNO}_{3}, \Delta]{\mathrm{KOH}} \mathrm{K}_{2} \mathrm{MnO}_{4}$
MCQ 104 Marks
The carbohydrates "Ribose" present in DNA, is
A. A pentose sugar
B. present in pyranose from
C. in "D" configuration
D. a reducing sugar, when free
E. in $\alpha$-anomeric form
Choose the correct answer from the options given below :
A. A pentose sugar
B. present in pyranose from
C. in "D" configuration
D. a reducing sugar, when free
E. in $\alpha$-anomeric form
Choose the correct answer from the options given below :
- ✓A, C and D Only
- BA, B and E Only
- CB, D and E Only
- DA, D and E Only
Answer
View full question & answer→Correct option: A.
A, C and D Only
(A)
Sol.
Sol.
MCQ 114 Marks
Which of the following Statements are NOT true about the periodic table?
A. The properties of elements are function of atomic weights.
B. The properties of elements are function of atomic numbers.
C. Elements having similar outer electronic configuration are arranged in same period.
D. An element's location reflects the quantum numbers of the last filled orbital.
E. The number of elements in a period is same as the number of atomic orbitals available in energy level that is being filled.
Choose the correct answer from the options given below:
A. The properties of elements are function of atomic weights.
B. The properties of elements are function of atomic numbers.
C. Elements having similar outer electronic configuration are arranged in same period.
D. An element's location reflects the quantum numbers of the last filled orbital.
E. The number of elements in a period is same as the number of atomic orbitals available in energy level that is being filled.
Choose the correct answer from the options given below:
- ✓A, C and E Only
- BD and E Only
- CA and E Only
- DB, C and E Only
Answer
View full question & answer→Correct option: A.
A, C and E Only
(A)
Sol. Properties of elements are periodic function of their atomic number. Elements having similar outer electronic configuration are arranged in same group. Number of elements in a period is not equal to number of atomic orbitals available in energy level that is being filled.
Hence, A, C & E are incorrect
Sol. Properties of elements are periodic function of their atomic number. Elements having similar outer electronic configuration are arranged in same group. Number of elements in a period is not equal to number of atomic orbitals available in energy level that is being filled.
Hence, A, C & E are incorrect
MCQ 124 Marks
Given below are two statements I and II.
Statement I : Dumas method is used for estimation of "Nitrogen" in an organic compound.
Statement II : Dumas method involves the formation of ammonium sulphate by heating the organic compound with conc $\mathrm{H}_{2} \mathrm{SO}_{4}$.
In the light of the above statements, choose the correct answer from the options given below
Statement I : Dumas method is used for estimation of "Nitrogen" in an organic compound.
Statement II : Dumas method involves the formation of ammonium sulphate by heating the organic compound with conc $\mathrm{H}_{2} \mathrm{SO}_{4}$.
In the light of the above statements, choose the correct answer from the options given below
- ABoth Statement I and Statement II are true
- BStatement I is false but Statement II is true
- CBoth Statement I and Statement II are false
- ✓Statement I is true but Statement II is false
Answer
View full question & answer→Correct option: D.
Statement I is true but Statement II is false
(D)
Sol. In Dumas method nitrogen present in organic compound is converted into $\mathrm{N}_{2}$ gas whose volumetric analysis gives the percentage of nitrogen atom in the organic compound.
Sol. In Dumas method nitrogen present in organic compound is converted into $\mathrm{N}_{2}$ gas whose volumetric analysis gives the percentage of nitrogen atom in the organic compound.
MCQ 134 Marks
Let us consider an endothermic reaction which is non-spontaneous at the freezing point of water. However, the reaction is spontaneous at boiling point of water. Choose the correct option.
- ✓Both $\Delta \mathrm{H}$ and $\Delta \mathrm{S}$ are (+ve)
- B$\Delta \mathrm{H}$ is $(-\mathrm{ve})$ but $\Delta \mathrm{S}$ is $(+\mathrm{ve})$
- C$\Delta \mathrm{H}$ is $(+\mathrm{ve})$ but $\Delta \mathrm{S}$ is $(-\mathrm{ve})$
- DBoth $\Delta \mathrm{H}$ and $\Delta \mathrm{S}$ are ( -ve )
Answer
View full question & answer→Correct option: A.
Both $\Delta \mathrm{H}$ and $\Delta \mathrm{S}$ are (+ve)
(A)
Sol. Reaction is spontaneous at relatively high temperature and non-spontaneous at low temperature $\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$
It is only possible when $\Delta \mathrm{H}$ and $\Delta \mathrm{S}$ both are positive.
Sol. Reaction is spontaneous at relatively high temperature and non-spontaneous at low temperature $\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$
It is only possible when $\Delta \mathrm{H}$ and $\Delta \mathrm{S}$ both are positive.
MCQ 144 Marks
Ksp for $\mathrm{Cr}(\mathrm{OH})_{3}$ is $1.6 \times 10^{-30}$. What is the molar solubility of this salt in water?
- ✓$\sqrt[4]{\frac{1.6 \times 10^{-30}}{27}}$
- B$\frac{1.8 \times 10^{-30}}{27}$
- C$\sqrt[5]{1.8 \times 10^{-30}}$
- D$\sqrt[2]{1.6 \times 10^{-30}}$
Answer
View full question & answer→Correct option: A.
$\sqrt[4]{\frac{1.6 \times 10^{-30}}{27}}$
(A)
Sol.
$
\mathrm{Cr}(\mathrm{OH})_{3(\mathrm{~s})} \rightleftharpoons \mathrm{Cr}_{(\mathrm{aq})}^{+3}+3 \mathrm{OH}_{(\mathrm{aq})}^{-}
$
At eq : $\quad \mathrm{s} \quad 3 \mathrm{~s}$
$\mathrm{K}_{\mathrm{sp}}=(\mathrm{s}) \cdot(3 \mathrm{~s})^{3}=27 \mathrm{~s}^{4}$
$27 \mathrm{~s}^{4}=1.6 \times 10^{-30}$
$\mathrm{s}=\left(\frac{1.6}{27} \times 10^{-30}\right)^{1 / 4}$
Sol.
$
\mathrm{Cr}(\mathrm{OH})_{3(\mathrm{~s})} \rightleftharpoons \mathrm{Cr}_{(\mathrm{aq})}^{+3}+3 \mathrm{OH}_{(\mathrm{aq})}^{-}
$
At eq : $\quad \mathrm{s} \quad 3 \mathrm{~s}$
$\mathrm{K}_{\mathrm{sp}}=(\mathrm{s}) \cdot(3 \mathrm{~s})^{3}=27 \mathrm{~s}^{4}$
$27 \mathrm{~s}^{4}=1.6 \times 10^{-30}$
$\mathrm{s}=\left(\frac{1.6}{27} \times 10^{-30}\right)^{1 / 4}$
MCQ 154 Marks
Which of the following ions is the strongest oxidizing agent ? (Atomic Number of $\mathrm{Ce}=58, \mathrm{Eu}=63, \mathrm{~Tb}=65, \mathrm{Lu}=71$ ]
- A$\mathrm{Lu}^{3+}$
- B$\mathrm{Eu}^{2+}$
- ✓$\mathrm{Tb}^{4+}$
- D$\mathrm{Ce}^{3+}$
Answer
View full question & answer→Correct option: C.
$\mathrm{Tb}^{4+}$
(C)
Sol. $Tb ^4$ is strongest oxidising agent
Sol. $Tb ^4$ is strongest oxidising agent
MCQ 164 Marks
Which of the following linear combination of atomic orbitals will lead to formation of molecular orbitals in homonuclear diatomic molecules [internuclear axis in z-direction] ?
A. $2 p_{z}$ and $2 p_{x}$
B. 2 s and $2 \mathrm{p}_{\mathrm{x}}$
C. $3 d_{x y}$ and $3 d_{x^{2}-y^{2}}$
D. 2 s and $2 \mathrm{p}_{z}$
E. $2 p_{z}$ and $3 d_{x^{2}-y^{2}}$
A. $2 p_{z}$ and $2 p_{x}$
B. 2 s and $2 \mathrm{p}_{\mathrm{x}}$
C. $3 d_{x y}$ and $3 d_{x^{2}-y^{2}}$
D. 2 s and $2 \mathrm{p}_{z}$
E. $2 p_{z}$ and $3 d_{x^{2}-y^{2}}$
- AE Only
- BA and B Only
- ✓D Only
- DC and D Only
Answer
View full question & answer→Correct option: C.
D Only
(C)
Question 174 Marks
Answer
View full question & answer→(B)
Sol. Carbocation intermediate is stabilised by $+\mathrm{I},+\mathrm{M}$ & hyperconjugation effect. Since in option 2 carbocation is in conjugation with stronger +M group $-\mathrm{OCH}_{3}$ hence it will be most stable.
Sol. Carbocation intermediate is stabilised by $+\mathrm{I},+\mathrm{M}$ & hyperconjugation effect. Since in option 2 carbocation is in conjugation with stronger +M group $-\mathrm{OCH}_{3}$ hence it will be most stable.
MCQ 184 Marks
One mole of the octahedral complex compound $\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}_{3}$ gives 3 moles of ions on dissolution in water. One mole of the same complex reacts with excess of $\mathrm{AgNO}_{3}$ solution to yield two moles of $\mathrm{AgCl}_{(\mathrm{s})}$. The structure of the complex is :
- ✓$\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2}$
- B$\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}\right] \cdot \mathrm{Cl}_{2} \cdot \mathrm{NH}_{3}$
- C$\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl} \cdot \mathrm{NH}_{3}$
- D$\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{3} \mathrm{Cl}_{3}\right] \cdot 2 \mathrm{NH}_{3}$
Answer
View full question & answer→Correct option: A.
$\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2}$
(A)
Sol.
$\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2} \rightarrow \underbrace{\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right]^{2+}(\text { aq. })+2 \mathrm{Cl}^{-}(\text {aq. })}_{3 \text { ions in water }}$
$\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2}$ (aq.) $+2 \mathrm{AgNO}_{3}$ (aq.) $\rightarrow$
$\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right]\left(\mathrm{NO}_{3}\right)_{2}$ (aq.) $+2 \mathrm{AgCl}(\mathrm{s})$
Sol.
$\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2} \rightarrow \underbrace{\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right]^{2+}(\text { aq. })+2 \mathrm{Cl}^{-}(\text {aq. })}_{3 \text { ions in water }}$
$\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2}$ (aq.) $+2 \mathrm{AgNO}_{3}$ (aq.) $\rightarrow$
$\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right]\left(\mathrm{NO}_{3}\right)_{2}$ (aq.) $+2 \mathrm{AgCl}(\mathrm{s})$
MCQ 194 Marks
Following are the four molecules "P", "Q", "R" and "S". Which one among the four molecules will react with $\mathrm{H}-\mathrm{Br}(\mathrm{aq})$ at the fastest rate ?
- AS
- ✓Q
- CR
- DP
Answer
View full question & answer→Correct option: B.
Q
(B)
MCQ 204 Marks
For the given cell
$
\mathrm{Fe}^{2+}(\mathrm{eq})+\mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{Ag}(\mathrm{~s})
$
The standard cell potential of the above reaction is Given :
$\mathrm{Ag}^{+}+\mathrm{e}^{-} \rightarrow \mathrm{Ag}$
$\mathrm{E}^{0}=\mathrm{xV}$
$\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Fe}$
$\mathrm{E}^{0}=\mathrm{yV}$
$\mathrm{Fe}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe}$
$\mathrm{E}^{0}=\mathrm{zV}$
$
\mathrm{Fe}^{2+}(\mathrm{eq})+\mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{Ag}(\mathrm{~s})
$
The standard cell potential of the above reaction is Given :
$\mathrm{Ag}^{+}+\mathrm{e}^{-} \rightarrow \mathrm{Ag}$
$\mathrm{E}^{0}=\mathrm{xV}$
$\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Fe}$
$\mathrm{E}^{0}=\mathrm{yV}$
$\mathrm{Fe}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe}$
$\mathrm{E}^{0}=\mathrm{zV}$
- A$x+y-z$
- ✓$x+2 y-3 z$
- C$y-2 x$
- D$x+2 y$
Answer
View full question & answer→Correct option: B.
$x+2 y-3 z$
(B)
$\Delta \mathrm{G}_{3}^{0}=\Delta \mathrm{G}_{1}^{0}+\Delta \mathrm{G}_{2}^{0}$
$-3 \mathrm{~F}(-\mathrm{z})=-2 \mathrm{~F}(-\mathrm{y})+\Delta \mathrm{G}_{2}{ }^{0}$
$\Delta \mathrm{G}_{2}^{0}=3 \mathrm{Fz}-2 \mathrm{Fy}$
Also $\Delta \mathrm{G}_{2}^{0}=-\mathrm{nFE}^{0} \mathrm{Fe}^{+2} / \mathrm{Fe}^{+3}$
$3 \mathrm{Fz}-2 \mathrm{Fy}=-1 \mathrm{~F}\left(\mathrm{E}_{\mathrm{Fe}^{+2} / \mathrm{Fe}^{+3}}^{0}\right)$
$\mathrm{E}_{\mathrm{Fe}^{+2} / \mathrm{Fe}^{+3}}^{0}=2 \mathrm{y}-3 \mathrm{z}$
$\mathrm{E}_{\text {Cell }}^{0}$ for reaction will be
$\mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{0}+\mathrm{E}_{\mathrm{Fe}^{+2} / \mathrm{Fe}^{+3}}^{0}$
$=x+2 y-3 z$
$\Delta \mathrm{G}_{3}^{0}=\Delta \mathrm{G}_{1}^{0}+\Delta \mathrm{G}_{2}^{0}$
$-3 \mathrm{~F}(-\mathrm{z})=-2 \mathrm{~F}(-\mathrm{y})+\Delta \mathrm{G}_{2}{ }^{0}$
$\Delta \mathrm{G}_{2}^{0}=3 \mathrm{Fz}-2 \mathrm{Fy}$
Also $\Delta \mathrm{G}_{2}^{0}=-\mathrm{nFE}^{0} \mathrm{Fe}^{+2} / \mathrm{Fe}^{+3}$
$3 \mathrm{Fz}-2 \mathrm{Fy}=-1 \mathrm{~F}\left(\mathrm{E}_{\mathrm{Fe}^{+2} / \mathrm{Fe}^{+3}}^{0}\right)$
$\mathrm{E}_{\mathrm{Fe}^{+2} / \mathrm{Fe}^{+3}}^{0}=2 \mathrm{y}-3 \mathrm{z}$
$\mathrm{E}_{\text {Cell }}^{0}$ for reaction will be
$\mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{0}+\mathrm{E}_{\mathrm{Fe}^{+2} / \mathrm{Fe}^{+3}}^{0}$
$=x+2 y-3 z$