Question 14 Marks
Which of the following compounds will behave as a reducing sugar in an aqueous KOH solution?
Questions
SECTION - A [CHEMISTY - MCQ]
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Question 24 Marks
$CH _3- CHCI - CH _2- CH _3$ has a chiral centre which one of the following represents its R - configuration?
Answer
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MCQ 34 Marks
When iodine reacts with sodium thiosulphate, the oxidation state of sulphur changes from +4 to __________ .
- A-2
- B+2
- C$+\frac{3}{2}$
- D$+\frac{5}{2}$
Answer
View full question & answer→D. $+\frac{5}{2}$
Explanation:
$I_2+2 \stackrel{+4}{S_2} O_{3(a q)}^{2-} \rightarrow \stackrel{+\frac{5}{S^2}}{S_4} O_{6(a q)}^{2-}+2 I_{(a q)}^{-}$
S has fractional oxidation state in the product.
Explanation:
$I_2+2 \stackrel{+4}{S_2} O_{3(a q)}^{2-} \rightarrow \stackrel{+\frac{5}{S^2}}{S_4} O_{6(a q)}^{2-}+2 I_{(a q)}^{-}$
S has fractional oxidation state in the product.
MCQ 44 Marks
Which techniques is based on the difference in the solubilities of the compound and the impurities in a suitable solvent?
- ACrystallisation
- BSublimation
- CCondensation
- DDistillation
Answer
View full question & answer→A. Crystallisation
Explanation:
Crystallisation process is based on the difference in the solubilities of the compound and the impurities in a suitable solvent.
The impure compound is dissolved in a solvent in which it is sparingly soluble at room temperature but appreciably soluble at higher temperature.
Explanation:
Crystallisation process is based on the difference in the solubilities of the compound and the impurities in a suitable solvent.
The impure compound is dissolved in a solvent in which it is sparingly soluble at room temperature but appreciably soluble at higher temperature.
MCQ 54 Marks
- A$A \longrightarrow CH _3 COOH , B\longrightarrow CH _3 COONa$
- B$A \longrightarrow NH _4 OH , B \longrightarrow NH _4 Cl$
- C$A \longrightarrow NH _4 OH , B \longrightarrow NaCl$
- D$A \longrightarrow KCl , B \longrightarrow NH _4 OH$
MCQ 64 Marks
Which of the following compounds can be prepared in good yield by Gabriel phthalimide synthesis?
- A$CH _3- CH _2- NHCH _3$
- B
- C
- D
MCQ 74 Marks
RNA and DNA are chiral molecules, their chirality is due to:
- Achiral bases
- Bchiral phosphate ester units
- CD-sugar component
- DL-sugar component
MCQ 84 Marks
Consider the given reaction, percentage yield of:
- AA $>$ C $>$ B
- BC $>$ A $>$ B
- CC $>$ B $>$ A
- DB $>$ C $>$ A
MCQ 94 Marks
Predict Y and Z.- AY = -Propylbenzene, Z = 2 Propylbenzaldehyde
- BY = Isopropylbenzene, Z = Benzoic acid
- CY = Isopropylbenzene, Z = Phenol
- DY = Propylphenol, Z = Benzophenone
MCQ 104 Marks
- A2-methylbut-1-en-1-ol
- Bpent-3-en-2-ol
- Cpent-4-en-2-ol
- Dpent-2-en-4-ol
Answer
View full question & answer→B. pent-3-en-2-ol
Explanation:
pent-3-en-2-ol
Explanation:
pent-3-en-2-ol
MCQ 114 Marks
Propene on hydroboration oxidation produces:
- A$CH _3 CH _2 CH _2 O$
- B$CH _3 CH _2 CHO$
- C$CH _3 CHOHCH _2 OH$
- D$CH _3 CHOHCH _3$
Answer
View full question & answer→A. $CH _3 CH _2 CH _2 O$
Explanation:
Anti-Markovnikov addition of $H _2 O$ on alkene.
Explanation:
Anti-Markovnikov addition of $H _2 O$ on alkene.
MCQ 124 Marks
The following statements are TRUE about crystal field theory except:
- AIt does not explain $\pi$ bonding in complexes.
- BThe satisfactory explanation is not provided for the fact that water is a stronger ligand than $OH ^{-}$.
- CPartial covalent nature of metal ligand bond is not explained.
- DIt considers s , p and d orbitals of the central metal.
Answer
View full question & answer→D.It considers s , p and d orbitals of the central metal.
Explanation:
In crystal field theory, only d orbitals of the central metal are considered. There is no explanation for s and p orbitals.
Explanation:
In crystal field theory, only d orbitals of the central metal are considered. There is no explanation for s and p orbitals.
MCQ 134 Marks
In the coordination compound $K _4\left[ Ni ( CN )_4\right]$, the oxidation state of nickel is:
- A-1
- B0
- C+2
- D+1
Answer
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MCQ 144 Marks
When neutral or faintly alkaline $KMnO _4$ is treated with potassium iodide, iodide ion is converted into $X . X$ is __________
- A$IO _3^{-}$
- B$IO ^{-}$
- C$I _2$
- D$IO _4^{-}$
Answer
View full question & answer→A. $IO _3^{-}$
Explanation:
In neutral or faintly alkaline $KMnO _4$ solution, iodide ion is converted into iodate.
$2 MnO_4^{-}+H_2 O+I^{-} \rightarrow 2 MnO_2+2 OH^{-}+IO_3^{-}$
Explanation:
In neutral or faintly alkaline $KMnO _4$ solution, iodide ion is converted into iodate.
$2 MnO_4^{-}+H_2 O+I^{-} \rightarrow 2 MnO_2+2 OH^{-}+IO_3^{-}$
MCQ 154 Marks
When conc. $H _2 SO _4$ was treated with $K _4\left[ Fe ( CN )_6\right]$, CO gas was evolved. By mistake, somebody used dilute $H _2 SO _4$ instead of conc. $H _2 SO _4$ then the gas evolved was:
- AHCN
- B$CO _2$
- C$N _2$
- DCO
Answer
View full question & answer→A. HCN
MCQ 164 Marks
If outermost E.C. of Ce is $4 f ^1 5 d^1 6 s^2$, then select the incorrectly matched outermost E.C. of its various ions in ground state:
- A$Ce ^{2+} \longrightarrow 4 f ^1 5 d^1$
- B$Ce ^{3+} \longrightarrow 4 f ^1$
- C$Ce ^{4+} \longrightarrow 4 f ^0$
- D$Ce ^{2+} \longrightarrow 4 f ^2$
Answer
View full question & answer→A. $Ce ^{2+} \longrightarrow 4 f ^1 5 d^1$
Explanation:
$Ce ^{2+} \longrightarrow 4 f ^2$
Explanation:
$Ce ^{2+} \longrightarrow 4 f ^2$
MCQ 174 Marks
For a first order reaction, $A \rightarrow P , t _{\frac{1}{2}}$ (half-life) is 10 days. The time required for $\frac{1}{4}^{\text {th }}$ conversion of $A$ (in days) is: (In $2=0.693$, In $3=1.1$ ).
- A3.2
- B2.5
- C5
- D4.1
Answer
View full question & answer→D. 4.1
Explanation:
The half life $t _{\frac{1}{2}}=10$ days The decay constant,
$K=\frac{0.693}{t_{\frac{1}{2}}}=\frac{0.693}{10 \text { days }}=0.0693 \text { days }^{-1}$
The time required for one fourth conversion
$\begin{array}{l}
t=\frac{2.303}{k} \log _{10} \frac{a}{a-x} \\
=\frac{2.303}{0.0693 day ^{-1}}\log _{10} \frac{1}{1-\left(\frac{1}{4}\right)}=4.1 \text { days }
\end{array}$
Explanation:
The half life $t _{\frac{1}{2}}=10$ days The decay constant,
$K=\frac{0.693}{t_{\frac{1}{2}}}=\frac{0.693}{10 \text { days }}=0.0693 \text { days }^{-1}$
The time required for one fourth conversion
$\begin{array}{l}
t=\frac{2.303}{k} \log _{10} \frac{a}{a-x} \\
=\frac{2.303}{0.0693 day ^{-1}}\log _{10} \frac{1}{1-\left(\frac{1}{4}\right)}=4.1 \text { days }
\end{array}$
MCQ 184 Marks
For a first order reaction $A \longrightarrow$ Product, the initial concentration of A is 0.1 M and after 40 minute it becomes 0.025 M . Calculate the rate of reaction at reactant concentration of 0.01 M .
- A$3.47 \times 10^{-5} M min ^{-1}$
- B$1.735 \times 10^{-6} M min ^{-1}$
- C$3.47 \times 10^{-4} M min ^{-1}$
- D$1.735 \times 10^{-4} M min ^{-1}$
Answer
View full question & answer→C. $3.47 \times 10^{-4} M min ^{-1}$
Explanation:
$\begin{array}{l} K =\frac{2.303}{40} \log \frac{0.1}{0.025} \\ \therefore K=0.03466 min^{-1} \\ \text { rate }= K \times 0.01=0.03466 \times 0.01=3.47 \times 10^{-4} M min ^{-1}\end{array}$
Explanation:
$\begin{array}{l} K =\frac{2.303}{40} \log \frac{0.1}{0.025} \\ \therefore K=0.03466 min^{-1} \\ \text { rate }= K \times 0.01=0.03466 \times 0.01=3.47 \times 10^{-4} M min ^{-1}\end{array}$
MCQ 194 Marks
Which of the following statements is incorrect for salt bridge?
- AIt maintains electrical neutrality of two half-cell solution.
- BIt prevents mixing of the electrolytic solutions.
- CIt completes the electrical circuit.
- DIt prevents the flow of current between the two half-cells.
Answer
View full question & answer→D. It prevents the flow of current between the two half-cells
MCQ 204 Marks
$AgNO _3$ (aq.) was added to an aqueous KCl solution gradually and the conductivity of the solution was measured. The plot of conductance $(\Lambda)$ versus the volume of $AgNO _3$ is:
- A(P)
- B(R)
- C(Q)
- D(S)
Answer
View full question & answer→D. (S)
Explanation:
Initial conductance $(\Lambda)$ of solution was due to $K _{(a q)}^{+}$and $Cl _{(a q)}^{-}$. On addition of $AgNO _3$, the reaction occurs as
$AgNO_3(aq)+KCl(aq) \rightarrow AgCl \downarrow+KNO_3$
Showing $AgNO _3$ as limiting reagent upto complete precipitation. The conductance up to precipitation shows horizontal linearly due to $K _{(a q)}^{+}$and (left $Cl ^{-}$formed $NO _3$ ) anions as ionic mobility of $NO _3$ and CP are almost same. After complete precipitation, further addition of $AgNO _3$, the conductance is observed due to $K ^{+}, NO _3$ ions and added $Ag ^{+}, NO _3$ ions thus a sudden increase is noticed.
Explanation:
Initial conductance $(\Lambda)$ of solution was due to $K _{(a q)}^{+}$and $Cl _{(a q)}^{-}$. On addition of $AgNO _3$, the reaction occurs as
$AgNO_3(aq)+KCl(aq) \rightarrow AgCl \downarrow+KNO_3$
Showing $AgNO _3$ as limiting reagent upto complete precipitation. The conductance up to precipitation shows horizontal linearly due to $K _{(a q)}^{+}$and (left $Cl ^{-}$formed $NO _3$ ) anions as ionic mobility of $NO _3$ and CP are almost same. After complete precipitation, further addition of $AgNO _3$, the conductance is observed due to $K ^{+}, NO _3$ ions and added $Ag ^{+}, NO _3$ ions thus a sudden increase is noticed.
MCQ 214 Marks
At $pH =1.2$ and 1 bar pressure, potential of hydrogen electrode at 298 K is __________ [Given : $\log 1.2=0.079$ ]
- A0.00 V
- B-0.295 V
- C-0.0708 V
- D-0.059 V
Answer
View full question & answer→C. $-0.0708 V$
Explanation:
For the given hydrogen electrode,
Pressure of $H _2=1$ bar and $pH =1.2$
Pressure of $H _2$ is 1 bar (or 1 atm ) and $\left[ H ^{+}\right] \neq 1 M$
$\begin{array}{l}
E_{\frac{H^{+}}{H_2}}=\frac{0.059}{2} \log _{10}\left[H^{+}\right]^2 \\
=0.059 \times \log _{10}\left[H^{+}\right] \\
=0.059 \times[-pH] \\
E_{H}=0.059 \times(-pH)=0.059 \times(-1.2)=-0.0708 V
\end{array}$
Explanation:
For the given hydrogen electrode,
Pressure of $H _2=1$ bar and $pH =1.2$
Pressure of $H _2$ is 1 bar (or 1 atm ) and $\left[ H ^{+}\right] \neq 1 M$
$\begin{array}{l}
E_{\frac{H^{+}}{H_2}}=\frac{0.059}{2} \log _{10}\left[H^{+}\right]^2 \\
=0.059 \times \log _{10}\left[H^{+}\right] \\
=0.059 \times[-pH] \\
E_{H}=0.059 \times(-pH)=0.059 \times(-1.2)=-0.0708 V
\end{array}$
MCQ 224 Marks
The freezing point of a 0.05 molal solution of a non-electrolyte in water is: $\left( K _{ f }=1.86\right.$ $K _{\text {molality }}{ }^{-1}$ )
- A$-0.093^{\circ} C$
- B$-1.86^{\circ} C$
- C$-0.93^{\circ} C$
- D$0.093^{\circ} C$
Answer
View full question & answer→A. $-0.093^{\circ} C$
Explanation:
$\begin{array}{l}\Delta T _{ f }=1.86 \times 0.05=0.093 \\ \therefore T_{ f }=0-0.093=-0.093^{\circ} C \end{array}$
Explanation:
$\begin{array}{l}\Delta T _{ f }=1.86 \times 0.05=0.093 \\ \therefore T_{ f }=0-0.093=-0.093^{\circ} C \end{array}$
MCQ 234 Marks
An unknown substance X contains $66.66$ $\%$ carbon, $3.73$ $\%$ hydrogen and $29.62 \%$ oxygen. Molar mass of X is the same as its empirical molar mass. If 3.15 grams of the unknown substance is dissolved in 25 grams of benzene, what is the freezing point of the resulting solution?
Normal freezing point of benzene is $5.50^{\circ} C$ and the molal freezing-point depression constant, $K _{ f }$, for benzene is 5.12 K $kg mol ^{-1}$.
Normal freezing point of benzene is $5.50^{\circ} C$ and the molal freezing-point depression constant, $K _{ f }$, for benzene is 5.12 K $kg mol ^{-1}$.
- A$-8.2{ }^{\circ} C$
- B$-11.9^{\circ} C$
- C$-2.9^{\circ} C$
- D$-6.5^{\circ} C$
Answer
View full question & answer→D. $-6.5^{\circ} C$
Explanation:
$\begin{array}{l} C _{\frac{5.5}{1.85}} H _{\frac{3.75}{1.85}} O _{\frac{1.85}{1.85}} \\ \therefore \text { Empirical formula }= C _3 H _2 O \\ \text { Molar mass }=54 g / mol \\ \text { Now, } T _{ f } \text { (solution) }= T _{ f (\text { solvent })}^0-\Delta T _{ f } \\ \Delta T _{ f }= m \cdot K _{ f }=\frac{3.15}{54 \times 0.025} \times 5.12 \\ \Delta T_{ f }=11.95^{\circ} C \\ \therefore T _{ f (\text { solution })}=5.5-11.95=-6.45 \approx-6.5^{\circ} C \end{array}$
Explanation:
| Element | $\%$ Composition | Moles |
| C | 66.66 | $66.66 g \times \frac{1 mol}{12 g}=5.5 mol$ |
| H | 3.73 | $3.73 g \times \frac{1 mol}{1.008 g}=3.7 mol$ |
| O | 29.62 | $29.62 g \times \frac{1 mol}{16 g}=1.85 mol$ |
MCQ 244 Marks
An unripe mango placed in a concentrated salt solution to prepare pickle, shrivels because:
- Ait gains water due to endo osmosis
- Bit gains water due to reverse osmosis
- Cit loses water due to exo osmosis
- Dit loses water due to reverse osmosis
Answer
View full question & answer→C. it Loses water due to exo osmosis
Explanation:
When unripe mango is dipped in saturated solution then it loses water due to exosmosis.
Explanation:
When unripe mango is dipped in saturated solution then it loses water due to exosmosis.
MCQ 254 Marks
An aqueous solution of a non-volatile solute show vapour pressure of 90.9 mm . If vapour pressure of pure water is 100 mm at the same temperature and $K_f$ for water is 1.86 K molality $^{-1}$, what will be depression in f.pt. of solution?
- A0.1034
- B10.34
- C1.034
- D0.206
Answer
View full question & answer→B. 10.34
MCQ 264 Marks
Consider the following reactions:
Calculate the ratio of number of $sp ^2$ hybridized C - atoms to the number of $sp ^3$ hybridized C - atoms in products B and C .
Calculate the ratio of number of $sp ^2$ hybridized C - atoms to the number of $sp ^3$ hybridized C - atoms in products B and C .
- AProduct B - Ratio $2: 3$, Product C - Ratio $3:2$
- BProduct B - Ratio $1:2$, Product C - Ratio $1:2$
- CProduct B - Ratio $2:1$, Product C - Ratio $2:3$
- DProduct B - Ratio $2:1$, Product C - Ratio $2:1$
Answer
View full question & answer→
MCQ 274 Marks
A will have configuration
- Aboth true
- Bnone of these
- C
- D
Answer
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MCQ 284 Marks
Which of the following ion is most stable?
- A
- B
- C$CH _3 \stackrel{+}{ C } HCH _2 CH _3$
- D$CH _3 CH _2 \stackrel{+}{ C } H _2$
Answer
View full question & answer→A.
MCQ 294 Marks
Graphite $\xrightarrow[(\text { oxidation })]{\text { Strong oxdising,agent }} X($ acid $) \xrightarrow[\text { Dehydration }]{\Delta} Y, Y$ is:
- A$C _3 O _2$
- B$CO _2$
- CCO
- D$C _{12} O _9$
MCQ 304 Marks
Bucky ball or Buckminster fullerene is:
- Ait has $sp ^2$-hybridised nature and resembles with a soccer ball
- Ban allotrope of carbon
- Cit is referred as C-60
- DAll of these
Answer
View full question & answer→D. All of these
MCQ 314 Marks
Which of the following reactions is an example of a redox reaction?
i. $XeF _4+ O _2 F_2 \rightarrow XeF _6+ O _2$
ii. $XeF _2+ PF _5 \rightarrow[ XeF ]^{+} PF _6^{-}$
iii. $XeF _6+ H _2 O \rightarrow XeOF _4+2 HF$
iv. $XeF _6+2 H _2 O \rightarrow XeO _2 F_2+4 HF$
i. $XeF _4+ O _2 F_2 \rightarrow XeF _6+ O _2$
ii. $XeF _2+ PF _5 \rightarrow[ XeF ]^{+} PF _6^{-}$
iii. $XeF _6+ H _2 O \rightarrow XeOF _4+2 HF$
iv. $XeF _6+2 H _2 O \rightarrow XeO _2 F_2+4 HF$
- Aiii and iv
- Bii and iii
- Conly i
- Div and i
Answer
View full question & answer→C. only i
Explanation:
$\begin{array}{l} XeF _4+ O _2 F_2 \rightarrow XeF _6+ O _2 \\ Xe ^{4+} \rightarrow Xe ^{6+}+2 e \\ 2 e +\left( O ^{+1}\right)_2 \rightarrow O_2^0\end{array}$
Explanation:
$\begin{array}{l} XeF _4+ O _2 F_2 \rightarrow XeF _6+ O _2 \\ Xe ^{4+} \rightarrow Xe ^{6+}+2 e \\ 2 e +\left( O ^{+1}\right)_2 \rightarrow O_2^0\end{array}$
MCQ 324 Marks
The mass of phosphoric acid $\left( H _3 PO _4\right)$ required to prepare 550 mL of 0.40 N solution
i. assuming complete neutralisation of acid
ii. assuming reduction to $HPO _3^{2-}$ are respectively:
i. assuming complete neutralisation of acid
ii. assuming reduction to $HPO _3^{2-}$ are respectively:
- A$10.8 g, 10.8 g$
- B$7.19 g, 7.19 g$
- C$10.8 g, 7.19 g$
- D$7.19 g, 10.8 g$
Answer
View full question & answer→D. $7.19 g, 10.8 g$
MCQ 334 Marks
Which of the following is not an example of redox reaction?
i. $CuO + H _2 \rightarrow Cu + H _2 O$
ii. $Fe _2 O _3+3 CO \rightarrow 2 Fe +3 CO _2$
iii. $2 K+ F _2 \rightarrow 2 KF$
iv. $BaCl _2+ H _2 SO _4 \rightarrow BaSO _4+2 HCI$
i. $CuO + H _2 \rightarrow Cu + H _2 O$
ii. $Fe _2 O _3+3 CO \rightarrow 2 Fe +3 CO _2$
iii. $2 K+ F _2 \rightarrow 2 KF$
iv. $BaCl _2+ H _2 SO _4 \rightarrow BaSO _4+2 HCI$
- Aii and iii
- Biv and i
- Ci and ii
- Diii and iv
Answer
View full question & answer→B. iv and i
Explanation:
Neither there is an oxidant nor reductant or none of the species shows the change in oxidation no.
Explanation:
Neither there is an oxidant nor reductant or none of the species shows the change in oxidation no.
MCQ 344 Marks
The total number of different kind of buffers obtained during the titration of $H _3 PO _4$ with NaOH are:
- A2
- B1
- Czero
- D3
Answer
View full question & answer→D. 3
Explanation:
$NaH _2 PO _4+ H _3 PO _4 ; NaH _2 PO _4+ Na _2 HPO _4 ; Na _2 HPO _4+ Na _3 PO _4$
Explanation:
$NaH _2 PO _4+ H _3 PO _4 ; NaH _2 PO _4+ Na _2 HPO _4 ; Na _2 HPO _4+ Na _3 PO _4$
MCQ 354 Marks
For the reaction $H_2(g)+I_2(g) \rightarrow 2 H I(g)$ the change in enthalpy $(\Delta H)$ will be
- A0
- B$=\Delta E$
- C$>\Delta E$
- D$<\Delta E$
Answer
View full question & answer→B. $=\Delta E$
Explanation:
$\begin{array}{l}
H_{2(g)}+I_{2(g)} \rightarrow 2 H I_{(2 g)} \\
\Delta n=N o .
\end{array}$
of gaseous product - no. of gaseous reactant
$\begin{array}{l}
=2-(1+1)=0 \\
A s \Delta H=\Delta E+\Delta n R T
\end{array}$
$\begin{array}{l}
\Rightarrow \Delta H=\Delta E+0 \times R T \\
\Rightarrow \Delta H=\Delta E
\end{array}$
Explanation:
$\begin{array}{l}
H_{2(g)}+I_{2(g)} \rightarrow 2 H I_{(2 g)} \\
\Delta n=N o .
\end{array}$
of gaseous product - no. of gaseous reactant
$\begin{array}{l}
=2-(1+1)=0 \\
A s \Delta H=\Delta E+\Delta n R T
\end{array}$
$\begin{array}{l}
\Rightarrow \Delta H=\Delta E+0 \times R T \\
\Rightarrow \Delta H=\Delta E
\end{array}$
MCQ 364 Marks
A mixture contains $n _1$ moles of monoatomic gas and $n _2$ moles of diatomic gas. If $\gamma=$ 1.50 for mixture, the ratio of $n_1: n_2$ is:
- A$2: 1$
- B$1: 1$
- C$1: 2$
- D$2: 3$
Answer
View full question & answer→B. 1:1
MCQ 374 Marks
The bond order in $NO ^{+}$molecule is __________ .
- A4
- B1
- C3
- D2
Answer
View full question & answer→C. 3
Explanation:
Electronic configuration of $NO ^{+}$molecule is
$\begin{array}{l}
NO^{+}: K K(\sigma 2 s)^2\left(\sigma^* 2 s\right)^2\left(\sigma 2 p_2\right)^2\left(\pi 2 p_x\right)^2 \\
\equiv\left(\pi 2 p_\nu\right)^2\left(\pi^* 2 p_x\right)^0
\end{array}$
KK represents closed K shell structure $(\sigma 1 s)^2\left(\sigma^* 1 s\right)^2$
Bond order of $NO ^{+}$molecule $=\frac{N_b-N_a}{2}=\frac{8-2}{2}=3$
Explanation:
Electronic configuration of $NO ^{+}$molecule is
$\begin{array}{l}
NO^{+}: K K(\sigma 2 s)^2\left(\sigma^* 2 s\right)^2\left(\sigma 2 p_2\right)^2\left(\pi 2 p_x\right)^2 \\
\equiv\left(\pi 2 p_\nu\right)^2\left(\pi^* 2 p_x\right)^0
\end{array}$
KK represents closed K shell structure $(\sigma 1 s)^2\left(\sigma^* 1 s\right)^2$
Bond order of $NO ^{+}$molecule $=\frac{N_b-N_a}{2}=\frac{8-2}{2}=3$
MCQ 384 Marks
What is the formal charge on the nitrogen atom in $HNO _3$ ?
- A0
- B+1
- C+3
- D+5
Answer
View full question & answer→B. +1
MCQ 394 Marks
Which of the following compounds readily decomposes on heating to give ionic compound with higher value of lattice energy as compare to corresponding reactant?
i. $Li _2 CO _3$
ii. $LiBF _4$
iii. $BeSO _4$
iv. $Na _2 O _2$
v. $BeCO _3$
i. $Li _2 CO _3$
ii. $LiBF _4$
iii. $BeSO _4$
iv. $Na _2 O _2$
v. $BeCO _3$
- AAll of these
- Bi, iii, iv, v
- Ci, iv, v
- Di, v
Answer
View full question & answer→A. All of these
Explanation :
$Li _2 CO _3 \xrightarrow{\Delta} Li _2 O + CO _2$
$LiBF _4 \xrightarrow{\Delta} LiF + BF _3$
$BeSO _4 \xrightarrow{\Delta} BeO + SO _2+\frac{1}{2} O _2$
$Na _2 O _2 \xrightarrow{\Delta} Na _2 O +\frac{1}{2} O _2$
$BeCO _3 \xrightarrow{\Delta} BeO + CO _2$
In all of these above reactions, the lattice energy of the product is higher as compared to the reactant.
Explanation :
$Li _2 CO _3 \xrightarrow{\Delta} Li _2 O + CO _2$
$LiBF _4 \xrightarrow{\Delta} LiF + BF _3$
$BeSO _4 \xrightarrow{\Delta} BeO + SO _2+\frac{1}{2} O _2$
$Na _2 O _2 \xrightarrow{\Delta} Na _2 O +\frac{1}{2} O _2$
$BeCO _3 \xrightarrow{\Delta} BeO + CO _2$
In all of these above reactions, the lattice energy of the product is higher as compared to the reactant.
MCQ 404 Marks
Which of following is not correctly matched?
A. d-block element: electronic configuration is $n s^{0-2}(n-1) d^{1-10}$
B. p-block element: electronic configuration is $ns ^{1-2} np ^{1-6}$
C. s-block elemen: electronic configuration is $ns { }^{1-2}$
D. Ce: f-block's first member.
A. d-block element: electronic configuration is $n s^{0-2}(n-1) d^{1-10}$
B. p-block element: electronic configuration is $ns ^{1-2} np ^{1-6}$
C. s-block elemen: electronic configuration is $ns { }^{1-2}$
D. Ce: f-block's first member.
- AD
- BB
- CA
- DC
Answer
View full question & answer→B. B
Explanation:
p-block element: electronic configuration is $n s^{1-2} n p^{1-6}$
Explanation:
p-block element: electronic configuration is $n s^{1-2} n p^{1-6}$
MCQ 414 Marks
If the wave number of light is $5 \times 10^6 m^{-1}$, then the frequency associated with this light is __________ .
- A$1.6 \times 10^{-2} sec$
- B$1.5 \times 10^{15} sec ^{-1}$
- C$1.5 \times 10^{15} sec$
- D$1.6 \times 10^{-2} sec ^{-1}$
Answer
View full question & answer→B. $1.5 \times 10^{15} sec ^{-1}$
Explanation:
$\begin{array}{l} v = c \bar{v} \\ =3 \times 10^8 ms^{-1} \times 5 \times 10^6 m^{-1} \\ v =1.5 \times 10^{15} sec ^{-1}\end{array}$
Explanation:
$\begin{array}{l} v = c \bar{v} \\ =3 \times 10^8 ms^{-1} \times 5 \times 10^6 m^{-1} \\ v =1.5 \times 10^{15} sec ^{-1}\end{array}$
MCQ 424 Marks
A certain orbital has no angular nodes and two radial nodes. The orbital is
- A3p
- B3s
- C2p
- D2s
Answer
View full question & answer→B. 3s
Explanation:
3s orbital has no angular nodes and two radial nodes.
Explanation:
| Orbit | Angular nodes $\ell$ | Radial nodes$(n-\ell-1)$ |
| 2p | $\ell=1$ | 2-1-1=0 |
| 3p | $\ell=1$ | 3-1-1=1 |
| 3s | $\ell=0$ | 3-0-1=2 |
| 2s | $\ell=0$ | 2-0-1=1 |
MCQ 434 Marks
The uncertainty in the momentum of an electron is $10^{-5} kg ms ^{-1}$. The uncertainty in it position will be:
- A$1.05 \times 10^{-26} m$
- B$5.25 \times 10^{-28} m$
- C$1.05 \times 10^{-28} m$
- D$5.27 \times 10^{-30} m$
Answer
View full question & answer→D. $5.27 \times 10^{-30} m$
Explanation:
Use $\Delta x \cdot \Delta p=\frac{h}{4 \pi}$
Explanation:
Use $\Delta x \cdot \Delta p=\frac{h}{4 \pi}$
MCQ 444 Marks
For the redox reaction,
$MnO_4^{-}+C_2 O_4^{2-}+H^{+} \longrightarrow Mn^{2+}+CO_2+H_2 O$
The correct coefficients of the reactants for the balanced reaction are
$MnO_4^{-}+C_2 O_4^{2-}+H^{+} \longrightarrow Mn^{2+}+CO_2+H_2 O$
The correct coefficients of the reactants for the balanced reaction are
- A$MnO _4^{-}=5, C _2 O _4^{2-}=16, H ^{+}=2$
- B$MnO _4^{-}=16, C _2 O _4^{2-}=5, H ^{+}=2$
- C$MnO _4^{-}=2, C _2 O _4^{2-}=5, H ^{+}=16$
- D$MnO _4^{-}=2, C _2 O _4^{2-}=16, H ^{+}=5$
Answer
View full question & answer→C. $MnO _4^{-}=2, C _2 O _4^{2-}=5, H ^{+}=16$
Explanation:
The balanced redox reaction is
$2 MnO_4^{-}+5 C_2 O_4^{2-}+16 H^{+} \longrightarrow 2 Mn^{2+}+10 CO_2+16 H_2 O$
Hence, the coefficients of reactants in balanced reaction are 2,5 and 16 respectively.
Explanation:
The balanced redox reaction is
$2 MnO_4^{-}+5 C_2 O_4^{2-}+16 H^{+} \longrightarrow 2 Mn^{2+}+10 CO_2+16 H_2 O$
Hence, the coefficients of reactants in balanced reaction are 2,5 and 16 respectively.
MCQ 454 Marks
Using the given data, calculate the average atomic mass of argon.
| Isotope | Atomic mass (amu) | Relative abundance (%) |
| 36 Ar | 35.97 | 0.337 |
| 38 Ar | 37.96 | 0.063 |
| 40 Ar | 39.96 | 99.6 |
- A39.94 amu
- B38.00 amu
- C38.50 amu
- D40 amu
Answer
View full question & answer→A. 39.94 amu
Explanation:
$\begin{array}{l}\text { Average atomic mass }=\frac{\text { Sum of (Isotopic mass } \times \text { its } \% \text { abundance ) }}{100} \\ =\frac{(35.97 \times 0.337)+(37.96 \times 0.063)+(39.96 \times 99.6)}{100} \\ =39.94 amu \end{array}$
Explanation:
$\begin{array}{l}\text { Average atomic mass }=\frac{\text { Sum of (Isotopic mass } \times \text { its } \% \text { abundance ) }}{100} \\ =\frac{(35.97 \times 0.337)+(37.96 \times 0.063)+(39.96 \times 99.6)}{100} \\ =39.94 amu \end{array}$