MCQ 14 Marks
In which of the following molecule/ion all the bonds are not equal?
- A$XeF _4$
- B$BF _4^{-}$
- ✓$BrCl _3$
- D$SF _6$
Answer
View full question & answer→Correct option: C.
$BrCl _3$
(C) $BrCl _3$
Questions
SECTION - A [CHEMISTY - MCQ]
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45 questions · 42 auto-graded MCQ + 3 self-marked written.
MCQ 24 Marks
Which of the following has a shape different from the other three species ?
- A$BF _4^{-}$
- B$SO _4{ }^{2-}$
- ✓$I _3^{-}$
- D$PH _4^{+}$
Answer
View full question & answer→Correct option: C.
$I _3^{-}$
(C) $I _3^{-}$
$I _3{ }^{-}$is linear, where as rest molecules are tetrahedral.
$I _3{ }^{-}$is linear, where as rest molecules are tetrahedral.
MCQ 34 Marks
Among $NaCl , BeCl _2, BCI _3$ and $CCl _4$, the covalent bond character follows the order
- ✓$NaCl < BeCl _2< BCl _3< CCl _4$
- B$NaCl < BeCl _2> BCl _3> CCl _4$
- C$NaCl > BeCl _2< BCl _3< CCl _4$
- D$NaCl > BeCl _2> BCl _3> CCl _4$
Answer
View full question & answer→Correct option: A.
$NaCl < BeCl _2< BCl _3< CCl _4$
(A) $NaCl < BeCl _2< BCl _3< CCl _4$
Among $NaCl , BeCl _2, BCl _3$ and $CCl _4$, the covalent bond character follows the order $NaCl < BeCl _2<$ $BCl _3< CCl _4$
Among $NaCl , BeCl _2, BCl _3$ and $CCl _4$, the covalent bond character follows the order $NaCl < BeCl _2<$ $BCl _3< CCl _4$
MCQ 44 Marks
$CH_3 CH_2 COOH \xrightarrow[\text { Red } P]{Cl_2}(A) \xrightarrow{\text { Alc. } KOH}(B)$ Compound $(B)$ is :
- A
- B
- ✓$CH _2= CH - COOH$
- D$ClCH _2 CH _2 COOH$
Answer
View full question & answer→Correct option: C.
$CH _2= CH - COOH$
(C) $ClCH _2 CH _2 COOH$
MCQ 54 Marks
Choose the correct statements :
i. Dipole moment of phenol is less than cyclo hexanol.
ii. Phenol is acidic in nature due to resonance.
iii. Boiling point of isobutyl alcohol is less than terteary butyl alcohol.
iv. Isobutyl alcohol is primary alcohol.
i. Dipole moment of phenol is less than cyclo hexanol.
ii. Phenol is acidic in nature due to resonance.
iii. Boiling point of isobutyl alcohol is less than terteary butyl alcohol.
iv. Isobutyl alcohol is primary alcohol.
- Ai, iii, iv
- Bii, iv
- ✓i, ii, iv
- Di, ii, iii, iv
Answer
View full question & answer→Correct option: C.
i, ii, iv
(C) i, ii, iv
In phenol negative charge of oxygen atom is destributed in resonance.
Phenol is acidic due to resonance
B.P. > B.P.
In phenol negative charge of oxygen atom is destributed in resonance.
Phenol is acidic due to resonance
B.P. > B.P.
MCQ 64 Marks
Correct order of given alkyl halide for $S _{ N } 1$ reaction :
(a) $\left( CH _3\right)_3 C - Cl$
(b) $\left( CH _3\right)_3 C - Br$
(c) $\left( CH _3\right)_3 C - I$
(a) $\left( CH _3\right)_3 C - Cl$
(b) $\left( CH _3\right)_3 C - Br$
(c) $\left( CH _3\right)_3 C - I$
- ✓c > b > a
- Bc < b < a
- Cc = b = a
- Dc > b < a
Answer
View full question & answer→Correct option: A.
c > b > a
(A) c > b > a
$S _{ N } 1$ reaction order :
$R-I>R-Br>R-Cl$
$S _{ N } 1$ reaction order :
$R-I>R-Br>R-Cl$
MCQ 74 Marks
Correct order of second ionization enthalpy are :
- A$Sc < V < Cr < Ti$
- B$V < Sc < Cr < Ti$
- C$Sc < Cr < V < Ti$
- ✓$Sc < Ti < V < Cr$
Answer
View full question & answer→Correct option: D.
$Sc < Ti < V < Cr$
(D) $Sc < Ti < V < Cr$
MCQ 84 Marks
Identify the reaction order from the following rate constants.
$k=3 \times 10^{-4} s^{-1}$
$k=3 \times 10^{-4} s^{-1}$
- AZero order reaction
- ✓First order reaction
- CSecond order reaction
- DNone of these
Answer
View full question & answer→Correct option: B.
First order reaction
(B) First order reaction
The unit of a first order rate constant is $s ^{-1}$ therefore $k =3 \times 10^{-4} s^{-1}$ represents a first order reaction
The unit of a first order rate constant is $s ^{-1}$ therefore $k =3 \times 10^{-4} s^{-1}$ represents a first order reaction
MCQ 94 Marks
For the reaction $R \rightarrow P$, the concentration of a reactant changes from 0.03 M to 0.02 M in 25 minutes. Calculate the average rate of reaction using units of time in seconds:
- A$1.67 \times 10^{-6}$
- B$3.67 \times 10^{-3}$
- C$5.67 \times 10^{-3}$
- ✓$6.67 \times 10^{-6}$
Answer
View full question & answer→Correct option: D.
$6.67 \times 10^{-6}$
(D) $6.67 \times 10^{-6}$
Average rate of reaction $=-\frac{\Delta[ R ]}{\Delta t}$
$\Rightarrow-\frac{[R]_2-[R]_1}{t_2-t_1}$
$\Rightarrow-\frac{0.02-0.03}{25}=4 \times 10^{-4} \text{ M min} ^{-1}$
Also, it can be expressed in seconds as:
$\begin{aligned} \text { Average rate of reaction } & =\frac{4 \times 10^{-4}}{60} \\ = & 6.67 \times 10^{-6}\end{aligned}$
Average rate of reaction $=-\frac{\Delta[ R ]}{\Delta t}$
$\Rightarrow-\frac{[R]_2-[R]_1}{t_2-t_1}$
$\Rightarrow-\frac{0.02-0.03}{25}=4 \times 10^{-4} \text{ M min} ^{-1}$
Also, it can be expressed in seconds as:
$\begin{aligned} \text { Average rate of reaction } & =\frac{4 \times 10^{-4}}{60} \\ = & 6.67 \times 10^{-6}\end{aligned}$
MCQ 104 Marks
Faraday’s law of electrolysis is related to the
- Aatomic number of the anion
- ✓equivalent weight of the electrolyte
- Catomic number of reactants.
- DNone of these
Answer
View full question & answer→Correct option: B.
equivalent weight of the electrolyte
(B) Equivalent weight of the electrolyte
MCQ 114 Marks
Why does the conductivity of a solution decreases with dilution because :
- Athe number of ions per unit volume decreases
- ✓the number of ions per unit volume decreases
- Cthe number of ions per unit volume decreases
- DNone of these
Answer
View full question & answer→Correct option: B.
the number of ions per unit volume decreases
(B) the number of ions per unit volume decreases
Conductivity of a solution is the conductance of ions present in a unit volume of the solution. On dilution, the number of ions per unit volume decreases. Hence the conductivity decreases.
Conductivity of a solution is the conductance of ions present in a unit volume of the solution. On dilution, the number of ions per unit volume decreases. Hence the conductivity decreases.
MCQ 124 Marks
A molal solution is one that contain 0.5 mole of a solute in :
- ✓500 g of solvent
- B500 ml of solvent
- C500 ml of solution
- D500 g of solution
Answer
View full question & answer→Correct option: A.
500 g of solvent
(A) 500 g of solvent
Molality $=\frac{\text { mole }}{\text { wt. of solvent in } Kg }$
Molality $=\frac{\text { mole }}{\text { wt. of solvent in } Kg }$
MCQ 134 Marks
At equilibrium which does not be zero
i. $E_{\text {cell }}$
ii. $\Delta G_{\text {cell }}$
iii. $\Delta G ^{\circ}{ }_{\text {cell }}$
i. $E_{\text {cell }}$
ii. $\Delta G_{\text {cell }}$
iii. $\Delta G ^{\circ}{ }_{\text {cell }}$
- Ai and ii
- Bii and iii
- ✓Only iii
- Di , ii and iii
Answer
View full question & answer→Correct option: C.
Only iii
(C) Only iii
$\Delta G^o$ cell
$\Delta G^o$ cell
MCQ 144 Marks
In the given reaction, Find out X, Y and Z respectively :
- A$X=C_3 H_6, Y=C_4 H_8, Z=C_6 H_{12}$
- B$X = C _4 H _8, Y= C _6 H _{12}, Z= C _3 H _6$
- C$X=C_6 H_{12}, Y=C_3 H_6, Z=C_4 H_8$
- ✓$X=C_6 H_{12}, Y=C_4 H_8, Z=C_3 H_6$
Answer
View full question & answer→Correct option: D.
$X=C_6 H_{12}, Y=C_4 H_8, Z=C_3 H_6$
(D) $X=C_6 H_{12}, Y=C_4 H_8, Z=C_3 H_6$
MCQ 154 Marks
The general combustion equation for any alkane is :
$C_n H_{2 n+2}+(x) O_2 \rightarrow nCO_2+(n+1) H_2 O$
Select the correct option for x :
$C_n H_{2 n+2}+(x) O_2 \rightarrow nCO_2+(n+1) H_2 O$
Select the correct option for x :
- A$\left(\frac{2 n+1}{2}\right)$
- B$\left(\frac{2 n+1}{4}\right)$
- ✓$\left(\frac{3 n+1}{2}\right)$
- D$\left(\frac{4 n+1}{4}\right)$
Answer
View full question & answer→Correct option: C.
$\left(\frac{3 n+1}{2}\right)$
C
MCQ 164 Marks
Sodium salt of which acid will be needed for the preparation of propane by decarboxylation :
- ✓Butanoic acid
- BPropanoic acid
- CPentanoic acid
- DEthanoic acid
Answer
View full question & answer→Correct option: A.
Butanoic acid
(A) Butanoic acid
$CH _3 CH _2 CH _2 COO ^{-} Na ^{+}+ NaOH \xrightarrow{ CaO } CH _3 CH _2 CH _3+ Na _2 CO _3$
$CH _3 CH _2 CH _2 COO ^{-} Na ^{+}+ NaOH \xrightarrow{ CaO } CH _3 CH _2 CH _3+ Na _2 CO _3$
MCQ 174 Marks
Match the following Column-I with Column-II :
| Column-। | Column-।I |
| (a) o-Ethylanisole | i. |
| (b) p-Nitroaniline | ii.  |
| (c) 2,3 - Dibromo-1- phenylpentane | iii.  |
| (d) 4-Ethyl-1-fluoro-2- nitrobenzene | iv.  |
- Aa-i, b-ii, c-iii, d-iv
- Ba-ii, b-i, c-iv, d-iii
- Ca-iii, b-iv, c-i, d-ii
- Da-iv, b-iii, c-ii, d-i
Answer
View full question & answer→ (A) a-i, b-ii, c-iii, d-iv
| Column-। | Column-।I |
| (a) o-Ethylanisole | i. |
| (b) p-Nitroaniline | ii. |
| (c) 2,3 - Dibromo-1- phenylpentane | iii. |
| (d) 4-Ethyl-1-fluoro-2- nitrobenzene | iv. |
MCQ 184 Marks
The class of compound “Arenes”, What is its example :
- ✓Benzene
- BBut-1-yne
- C1-Bromobutane
- DNone of these
Answer
View full question & answer→Correct option: A.
Benzene
(A) Benzene
The class of compound Arenes – Example : Benzene
The class of compound Arenes – Example : Benzene
MCQ 194 Marks
What is the Common or Trivial Names of given Organic Compounds :
$\left(H_3 C\right)_2 CHCH_3$
$\left(H_3 C\right)_2 CHCH_3$
- An-Butane
- BNeopentane
- ✓Isobutane
- Dn-Propyl alcohol
Answer
View full question & answer→Correct option: C.
Isobutane
(C) Isobutane
- $H _3 CCH _2 CH _2 CH _3-$ n-Butane
- $\left( H _3 C \right)_4 C -$ Neopentane
- $\left( H _3 C _2 CHCH _3-\right.$ Isobutane
- $H _3 CCH _2 CH _2 OH -$ n-Propyl alcohol
- $H _3 CCH _2 CH _2 CH _3-$ n-Butane
- $\left( H _3 C \right)_4 C -$ Neopentane
- $\left( H _3 C _2 CHCH _3-\right.$ Isobutane
- $H _3 CCH _2 CH _2 OH -$ n-Propyl alcohol
MCQ 204 Marks
Boiling point of water at 750 mm Hg is $99.63^{\circ} C$. How much sucrose is to be added to 500 g of water such that it boils at $100^{\circ} C$. [Given $K _{ b }=0.52$ and molecuar weight of sucrose $=342$]
- ✓121.67 g
- B101.67 g
- C12.167 g
- D111.67 g
Answer
View full question & answer→Correct option: A.
121.67 g
A
MCQ 214 Marks
Which statement is correct :
A. Oxidation : Loss of electron(s) by any species.
B. Reduction : Gain of electron(s) by any species.
C. Oxidising agent : Donor of electron(s).
D. Reducing agent : Acceptor of electron(s).
A. Oxidation : Loss of electron(s) by any species.
B. Reduction : Gain of electron(s) by any species.
C. Oxidising agent : Donor of electron(s).
D. Reducing agent : Acceptor of electron(s).
- AA and D
- BC and D
- CB and C
- ✓A and B
Answer
View full question & answer→Correct option: D.
A and B
(D) A and B
- Oxidation : Loss of electron(s) by any species.
- Reduction : Gain of electron(s) by any species.
- Oxidising agent : Acceptor of electron(s).
- Reducing agent : Donor of electron(s).
- Oxidation : Loss of electron(s) by any species.
- Reduction : Gain of electron(s) by any species.
- Oxidising agent : Acceptor of electron(s).
- Reducing agent : Donor of electron(s).
MCQ 224 Marks
The boiling point of benzene is 353.23 K . When 1.80 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. Kb for benzene is $2.53 \text{ K kg mol} ^{-1}$ :
- A$18 \text{g mol} ^{-1}$
- B$28 \text{g mol} ^{-1}$
- ✓$58 \text{g mol} ^{-1}$
- D$78 \text{g mol} ^{-1}$
Answer
View full question & answer→Correct option: C.
$58 \text{g mol} ^{-1}$
(C) $58 \text{g mol} ^{-1}$
The elevation $\left(\Delta T_b\right)$ in the boiling point $=354.11 K-353.23 K=0.88 K$
Substituting these values in expression (2.33) we get
$\begin{array}{l} M _2=\frac{2.53 \text{ K kg mol} ^{-1} \times 1.8 \text{ g } \times 1000 \text{ g kg} ^{-1}}{0.88 \text{ K } \times 90 \text{ g }} \\ =58 \text{ g mol} ^{-1}\end{array}$
Therefore, molar mass of the solute,
$M_2=58 \text{ g mol}^{-1}$
The elevation $\left(\Delta T_b\right)$ in the boiling point $=354.11 K-353.23 K=0.88 K$
Substituting these values in expression (2.33) we get
$\begin{array}{l} M _2=\frac{2.53 \text{ K kg mol} ^{-1} \times 1.8 \text{ g } \times 1000 \text{ g kg} ^{-1}}{0.88 \text{ K } \times 90 \text{ g }} \\ =58 \text{ g mol} ^{-1}\end{array}$
Therefore, molar mass of the solute,
$M_2=58 \text{ g mol}^{-1}$
MCQ 234 Marks
The $NH _3- Co - NH _3$ bond angle values in a fac- $\left[ Co \left( NH _3\right)_3 Cl _3\right]$ complex is/are
- A$180^{\circ}$
- ✓$90^{\circ}$
- C$90^{\circ}$ and $180^{\circ}$
- D$90^{\circ}$ and $120^{\circ}$
Answer
View full question & answer→Correct option: B.
$90^{\circ}$
(B) $90^{\circ}$
MCQ 244 Marks
Match List - 1 with List - II:
Choose the correct answer from the options given below:
| List-I (Complexes) | List-II (Types) |
| (a) $\left[ Co \left( NH _3\right)_5 CN \right] Cl _2$ | (i) ionisation isomerism |
| (b) $\left[ Cr \left( NH _3\right)_6\right]\left[ Co ( CN )_6\right]$ | (ii) coordination isomerism |
| (c) $\left[ Co \left( NH _3\right)_5\left( SO _4\right)\right] Br$ | (iii) linkage isomerism |
- A(a)-(i), (b)-(iii), (c)-(ii)
- B(a)-(iii), (b)- (i), (c)-(ii)
- C(a)-(ii), (b)-(iii), (c)-(i)
- ✓(a)-(iii), (b)-(ii), (c)-(i)
Answer
View full question & answer→Correct option: D.
(a)-(iii), (b)-(ii), (c)-(i)
(D) (a)-(iii), (b)-(ii), (c)-(i)
(a)-(iii), (b)-(ii), (c)-(i)
| List-I (Complexes) | List-II (Types) |
| (a) $\left[ Co \left( NH _3\right)_5 CN \right] Cl _2$ | (i) linkage isomerism |
| (b) $\left[ Cr \left( NH _3\right)_6\right]\left[ Co ( CN )_6\right]$ | (ii) coordination isomerism |
| (c) $\left[ Co \left( NH _3\right)_5\left( SO _4\right)\right] Br$ | (iii) linkage isomerism |
MCQ 254 Marks
Given below are two statements :
Statement I: $Ce ^{4+}$ Diamagnetic in nature
Statement II: $Ce ^{4+}$ act as oxidising agent.
In the light of the above statements, choose the most appropriate answer from the options given below :
Statement I: $Ce ^{4+}$ Diamagnetic in nature
Statement II: $Ce ^{4+}$ act as oxidising agent.
In the light of the above statements, choose the most appropriate answer from the options given below :
- ABoth Statement I and Statement II are incorrect
- BStatement I is correct but Statement II is incorrect
- CStatement I is incorrect but Statement II is correct
- ✓Both Statement I and Statement II are correct.
Answer
View full question & answer→Correct option: D.
Both Statement I and Statement II are correct.
(D) Both Statement I and Statement II are correct.
- $Ce ^{4+}$ Diamagnetic in nature
- $Ce ^{4+}$ act as oxidising agent.
- $Ce ^{4+}$ Diamagnetic in nature
- $Ce ^{4+}$ act as oxidising agent.
MCQ 264 Marks
What is the bond lengths in common molecules O2 (O = O) :
- A92
- B109
- C121
- D127
Answer
View full question & answer→
Question 274 Marks
Name according to IUPAC nomenclature with IUPAC Official Name is incorrect :
| Name according to IUPAC nomenclature | IUPAC offficial Name |
| (A) Ununoctium | Oganesson |
| (B) Ununnillium | Darmstadtium |
| (C) Unniltrium | Lawrencium |
| (D) Unnilseptium | Dubnium |
Answer
View full question & answer→(D) Unnilseptium – Dubnium
- Unnilseptium - Bohrium
Atomic no. - 107
Symbol - Uns
- Unnilpentium - Dubnium
Atomic no. - 105
Symbol - Unp
- Unnilseptium - Bohrium
Atomic no. - 107
Symbol - Uns
- Unnilpentium - Dubnium
Atomic no. - 105
Symbol - Unp
MCQ 284 Marks
Calculate the frequency of yellow radiation having wavelength 5800 Å.
- A$1.172 \times 10^{14} s^{-1}$
- B$2.172 \times 10^{10} s^{-1}$
- C$3.172 \times 10^{10} s^{-1}$
- ✓$5.172 \times 10^{14} s^{-1}$
Answer
View full question & answer→Correct option: D.
$5.172 \times 10^{14} s^{-1}$
(D) $5.172 \times 10^{14} s^{-1}$
Calculation of the frequency (v)
$\bar{v}=\frac{c}{\lambda}=\frac{3 \times 10^8 m s ^{-1}}{5800 \times 10^{-10} m}$
$=5.172 \times 10^{14} s^{-1}$
Calculation of the frequency (v)
$\bar{v}=\frac{c}{\lambda}=\frac{3 \times 10^8 m s ^{-1}}{5800 \times 10^{-10} m}$
$=5.172 \times 10^{14} s^{-1}$
MCQ 294 Marks
Which of the following amines does not react with Hinsberg’s reagent :
- A$CH _3 CH _2- NH _2$
- B$CH _3- NH - CH _3$
- ✓$\left( CH _3 CH _2\right)_3 N$
- DAll of these
Answer
View full question & answer→Correct option: C.
$\left( CH _3 CH _2\right)_3 N$
(C) $\left( CH _3 CH _2\right)_3 N$
MCQ 304 Marks
In the reactions given below, identify the element undergoing oxidation and reduction:
$3 Fe_3 O_4(s)+8 Al(s) \rightarrow 9 Fe(s)+4 Al_2 O_3(s)$
$3 Fe_3 O_4(s)+8 Al(s) \rightarrow 9 Fe(s)+4 Al_2 O_3(s)$
- AAI, O
- BAI, Fe
- ✓Fe, O
- DNone of these
Answer
View full question & answer→Correct option: C.
Fe, O
(C) AI, Fe
Aluminium is oxidised because oxygen is added to it. Ferrous ferric oxide redox reactions $\left( Fe _3 O _4\right)$ is reduced because oxygen has been removed from it.
Aluminium is oxidised because oxygen is added to it. Ferrous ferric oxide redox reactions $\left( Fe _3 O _4\right)$ is reduced because oxygen has been removed from it.
MCQ 314 Marks
The oxidation of toluene to benzaldehyde by chromyl chloride is called :
- ✓Etard reaction
- BRiemer-Tiemann reaction
- CWurtz reaction
- DNone of these
Answer
View full question & answer→Correct option: A.
Etard reaction
A
MCQ 324 Marks
Which one is the commercial method for the prepration of benzaldehyde.
(a) Rosenmund reduction
(b) Etard reaction
(c) Stephen reaction
(d) Side chain chlorination, followed by hydrolysis.
(a) Rosenmund reduction
(b) Etard reaction
(c) Stephen reaction
(d) Side chain chlorination, followed by hydrolysis.
- Aa, b
- Bb, c
- Conly c
- ✓only d
Answer
View full question & answer→Correct option: D.
only d
(D) only d
Side chain chlorination, followed by hydrolysis.
Side chain chlorination, followed by hydrolysis.
MCQ 334 Marks
Oil of Eranda and seeds of Mahua plant and calcium carbonate were used for .
- Amaking medicine
- Bmaking vegetable
- Cmaking ornaments
- ✓making soap
Answer
View full question & answer→Correct option: D.
making soap
(D) making soap
Oil of Eranda and seeds of Mahua plant and calcium carbonate were used for making soap.
Oil of Eranda and seeds of Mahua plant and calcium carbonate were used for making soap.
MCQ 344 Marks
____________ discovered mercury sulphide :
- ✓Chakrapani
- BRsarnavam
- CRigveda
- DNone of these
Answer
View full question & answer→Correct option: A.
Chakrapani
(A) Chakrapani
Chakrapani discovered mercury sulphide.
Chakrapani discovered mercury sulphide.
MCQ 354 Marks
At equilibrium, the concentrations of $N _2=3.0 \times 10^{-3}$ $M , O _2=4.2 \times 10^{-3} M$ and $NO =2.8 \times 10^{-3} M$ in a sealed vessel at 800 K . What will be $K _{ c }$ for the reaction
$N_2(g)+O_2(g) \rightleftharpoons 2 NO(g)$
$N_2(g)+O_2(g) \rightleftharpoons 2 NO(g)$
- A1.622
- B6.22
- ✓0.622
- D0.222
Answer
View full question & answer→Correct option: C.
0.622
(C) 0.622
For the reaction equilibrium constant, $K _{ c }$ can be written as
$K _{ C }=\frac{[ NO ]^2}{\left[N_2\right]\left[ O _2\right]}$
$K _{ C }=\frac{\left(2.8 \times 10^{-3} M \right)^2}{\left(3.0 \times 10^{-3} M \right)\left(4.2 \times 10^{-3} M \right)}$
$=0.622$
For the reaction equilibrium constant, $K _{ c }$ can be written as
$K _{ C }=\frac{[ NO ]^2}{\left[N_2\right]\left[ O _2\right]}$
$K _{ C }=\frac{\left(2.8 \times 10^{-3} M \right)^2}{\left(3.0 \times 10^{-3} M \right)\left(4.2 \times 10^{-3} M \right)}$
$=0.622$
MCQ 364 Marks
Choose the correct answer. A thermodynamic state function is a quantity
- Aused to determine heat changes
- ✓whose value is independent of path
- Cused to determine pressure volume work
- Dwhose value depends on temperature only
Answer
View full question & answer→Correct option: B.
whose value is independent of path
(B) whose value is independent of path
In thermodynamics, state function is a property whose value depends only on the current state of the system rather than on the path taken to reach that specific value.
In thermodynamics, state function is a property whose value depends only on the current state of the system rather than on the path taken to reach that specific value.
MCQ 374 Marks
Identify the correct statements :
A. $F _2$ and $O _2{ }^{2-}$ have bond order 2.
B. $N _2, CO$ and $NO ^{+}$have bond order 3 .
A. $F _2$ and $O _2{ }^{2-}$ have bond order 2.
B. $N _2, CO$ and $NO ^{+}$have bond order 3 .
- AOnly A
- ✓Only B
- CA and B
- DNone of these
Answer
View full question & answer→Correct option: B.
Only B
(B) Only B
A. Isoelectronic molecules and ions have identical bond orders; for example, $F _2$ and $O _2{ }^{2-}$ have bond order 1.
B. $N _2, CO$ and $NO ^{+}$have bond order 3 .
A. Isoelectronic molecules and ions have identical bond orders; for example, $F _2$ and $O _2{ }^{2-}$ have bond order 1.
B. $N _2, CO$ and $NO ^{+}$have bond order 3 .
MCQ 384 Marks
The oxidation state and coordination number of Al in $\left[ AlCl \left( H _2 O \right)_5\right]^{2+}$ are :
- A6, +2
- ✓+3 , 6
- C+5, 6
- DNone of these
Answer
View full question & answer→Correct option: B.
+3 , 6
(B) +3, 6
The oxidation state of Al is +3 and The covalency is 6 .
The oxidation state of Al is +3 and The covalency is 6 .
MCQ 394 Marks
The correct order of electron affinity of B, C, Ne, O is :
- A$O > C > Ne > B$
- B$B > Ne > C > O$
- ✓$O > C > B > Ne$
- D$O > B > C > Ne$
Answer
View full question & answer→Correct option: C.
$O > C > B > Ne$
(C) $O > C > B > Ne$
Order of electron affinity
$O > C > B > Ne$
Order of electron affinity
$O > C > B > Ne$
MCQ 404 Marks
If pH of 0.1M acetic acid is 3.0, then dissociation constant of this acid will be
- A$10^{-4}$
- ✓$10^{-5}$
- C$10^{-3}$
- D$10^{-8}$
Answer
View full question & answer→Correct option: B.
$10^{-5}$
(B) $10^{-5}$
${\left[ H ^{+}\right] 10^{-3}, pH =3 \quad C =10^{-1}}$
${\left[ H ^{+}\right]= C \cdot x }$
$10^{-3}=10^{-1} \cdot x$
$x =\frac{10^{-3}}{10^{-1}}=10^{-2}, K_{ a }= Cx { }^2$
$K_{ a }=C x^2=10^{-1} \times 10^{-2} \times 10^{-2}=10^{-5}$
${\left[ H ^{+}\right] 10^{-3}, pH =3 \quad C =10^{-1}}$
${\left[ H ^{+}\right]= C \cdot x }$
$10^{-3}=10^{-1} \cdot x$
$x =\frac{10^{-3}}{10^{-1}}=10^{-2}, K_{ a }= Cx { }^2$
$K_{ a }=C x^2=10^{-1} \times 10^{-2} \times 10^{-2}=10^{-5}$
MCQ 414 Marks
A reaction $X+Y \rightarrow Z+A+q$ is found to have a positive entropy change, the reaction will be :
- APossible at high temperature
- BPossible only at low temperature
- CNot possible at any temperature
- ✓Possible at any temperature
Answer
View full question & answer→Correct option: D.
Possible at any temperature
(D) Possible at any temperature
$ \begin{array}{l} \Delta H=-v e \\ \Delta S=+v e \\ \Delta G=\Delta H-T \Delta S \end{array} $
$\Delta G$ is -ve at all temperaters.
$ \begin{array}{l} \Delta H=-v e \\ \Delta S=+v e \\ \Delta G=\Delta H-T \Delta S \end{array} $
$\Delta G$ is -ve at all temperaters.
MCQ 424 Marks
For which of these reactions will there be $\Delta$S negative :
- ✓$NH _3(g) \rightarrow NH _3( l )$
- B$H _2(g)+ I _2(g) \rightarrow 2 HI ( g )$
- C$CaCO _3$ (s) $\rightarrow CaO$ (s) $+ CO _2$ (g)
- DAll of these
Answer
View full question & answer→Correct option: A.
$NH _3(g) \rightarrow NH _3( l )$
(A) $NH _3(g) \rightarrow NH _3( l )$
Gas changes to liquid so entropy decreases.
Gas changes to liquid so entropy decreases.
MCQ 434 Marks
$O - O$ bond length order in $O _2^{-2}, O _2^{-}, O _3$ and $O _2$ :
- A$O _2^{-2}< O _2^{-}< O _3< O _2$
- B$O _2^{-2}> O _2^{-}> O _3> O _2$
- ✓$O _2^{-2}> O _2^{-}= O _3> O _2$
- D$O _2^{-2}< O _2^{-}= O _3< O _2$
Answer
View full question & answer→Correct option: C.
$O _2^{-2}> O _2^{-}= O _3> O _2$
(C) $O _2^{-2}> O _2^{-}= O _3> O _2$
Bond order $\propto\frac{1}{\text{ Bond lenght}} $
$\begin{array}{l} B . O = O _2^{-2}> O _2^{-}= O _3> O _2 \\ ( O - O )\end{array}$
Bond order $\propto\frac{1}{\text{ Bond lenght}} $
$\begin{array}{l} B . O = O _2^{-2}> O _2^{-}= O _3> O _2 \\ ( O - O )\end{array}$
MCQ 444 Marks
In which conversion, bond length decreases :
- A$N _2 \rightarrow N_2{ }^{+}$
- B$N _2 \rightarrow N_2^{-}$
- ✓$O _2 \rightarrow O _2^{+}$
- D$O _2 \rightarrow O _2^{-}$
Answer
View full question & answer→Correct option: C.
$O _2 \rightarrow O _2^{+}$
(C) $O _2 \rightarrow O _2^{+}$
Bond length $\propto 1 /$ Bond order
Bond length $\propto 1 /$ Bond order
MCQ 454 Marks
Given below are two statements :
Statement I:
Both $H _3 O ^{+}$and $NH _3$ are pyramidal molecules.
Statement II:
Both $H _3 O ^{+}$and $NH _3$ have $sp ^3$ hybridisation of the central atom.
In the light of the above statements, choose the most appropriate answer from the options given below :
Statement I:
Both $H _3 O ^{+}$and $NH _3$ are pyramidal molecules.
Statement II:
Both $H _3 O ^{+}$and $NH _3$ have $sp ^3$ hybridisation of the central atom.
In the light of the above statements, choose the most appropriate answer from the options given below :
- ABoth Statement I and Statement II are incorrect
- BStatement I is correct but Statement II is incorrect
- CStatement I is incorrect but Statement II is correct
- ✓Both Statement I and Statement II are correct.
Answer
View full question & answer→Correct option: D.
Both Statement I and Statement II are correct.
(D) Both Statement I and Statement II are correct.
- Both $H _3 O ^{+}$and $NH _3$ are pyramidal molecules.
- Both $H _3 O ^{+}$and $NH _3$ have $sp ^3$ hybridisation of the central atom.
- Both $H _3 O ^{+}$and $NH _3$ are pyramidal molecules.
- Both $H _3 O ^{+}$and $NH _3$ have $sp ^3$ hybridisation of the central atom.