Question 14 Marks
Which of the following compounds is most reactive towards SN2 reaction.
Answer
View full question & answer→A
Questions
SECTION - A [CHEMISTY - MCQ]
Take a timed test
45 questions · 38 auto-graded MCQ + 7 self-marked written.
MCQ 24 Marks
Which of the following is correct w.r.t. acidic character:
- ✓$CH _3 COOH > C _6 H _5 OH > EtOH$
- B$C _6 H _5 OH > CH _3 COOH > EtOH$
- C$EtOH > CH _3 COOH > C _6 H _5 OH$
- D$EtOH > C _6 H _5 OH > CH _3 COOH$
Answer
View full question & answer→Correct option: A.
$CH _3 COOH > C _6 H _5 OH > EtOH$
(A) $CH _3 COOH > C _6 H _5 OH > EtOH$
Order of acidic strength is
Carboxylic acids > Phenols > Alcohols
Order of acidic strength is
Carboxylic acids > Phenols > Alcohols
MCQ 34 Marks
Catalyst used in the given reaction is
- APd
- BMo2O3
- ✓CuCl2
- DZnO-Cr2O3
Answer
View full question & answer→Correct option: C.
CuCl2
(C) ZnO-Cr2O3
MCQ 44 Marks
An organic compound (X) on reaction with ethanoic anhydride gives N-phenyl ethanamide. Correct statements about the compound (X) are
I. X on reaction with chloroform and ethanolic potassium hydroxide forms isocyanide
II. X on reaction with nitrous acid at low temperature forms benzene diazonium salt
III. X is more basic than pyridine
I. X on reaction with chloroform and ethanolic potassium hydroxide forms isocyanide
II. X on reaction with nitrous acid at low temperature forms benzene diazonium salt
III. X is more basic than pyridine
- AI and III only
- ✓I and II only
- CII and III only
- DAll I, II and III
Answer
View full question & answer→Correct option: B.
I and II only
(B) I and II only
MCQ 54 Marks
Which among the following is not true for a cyclic process:
- ✓W = 0
- B$\Delta U=0$
- C$\Delta H=0$
- D$\Delta S=0$
Answer
View full question & answer→Correct option: A.
W = 0
(A) W = 0
Work is a path function, therefore will not be zero for a cyclic process whereas rest all are state functions
Work is a path function, therefore will not be zero for a cyclic process whereas rest all are state functions
Question 64 Marks
The major product of the following reaction is
Answer
View full question & answer→(B)
MCQ 74 Marks
Strongest acid among the following is
- ✓HCOOH
- BC6H5COOH
- CCH3COOH
- DCH2CH2COOH
Answer
View full question & answer→Correct option: A.
HCOOH
(A) HCOOH
Acidic strength increases as pKa decreases. HCOOH has least pKa value so it will be strongest acid among the following.
Acidic strength increases as pKa decreases. HCOOH has least pKa value so it will be strongest acid among the following.
Question 84 Marks
Consider the following reaction sequence
Answer
View full question & answer→(B)
MCQ 94 Marks
Statement -I: Rate of reaction for end endothermic reaction increases with increases in temperature.
Statement-II: Rate of reaction for exothermic reaction decreases with increase in temperature.
In the light of above statements, choose the most appropriate answer from the options given below.
Statement-II: Rate of reaction for exothermic reaction decreases with increase in temperature.
In the light of above statements, choose the most appropriate answer from the options given below.
- AStatement I is correct but statement II is incorrect
- BStatement I is incorrect but statement II is correct
- CBoth statement I and statement II are incorrect
- ✓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
Rate of both endothermic and exothermic reaction increase with increase in temperature.
Rate of both endothermic and exothermic reaction increase with increase in temperature.
MCQ 104 Marks
The equivalent weight of $Cr _2\left( SO _4\right)_3[mol, wt = M ]$ in the following reaction is $Cr _2\left( SO _4\right)_3+ H _2 O _2+ NaH \rightarrow Na _2 CrO _4+ Na _2 SO _4+ H _2 O$
- AM/3
- BM/4
- CM
- DM/6
MCQ 114 Marks
Which of the following compounds can be best prepared by Wurtz reaction?
- An-pentane
- BIsopentane
- CIsobutane
- ✓n-butane
Answer
View full question & answer→Correct option: D.
n-butane
(D) n-butane
Wurtz reaction is useful to prepare symmetrical alkanes with even number of carbon atoms.
Wurtz reaction is useful to prepare symmetrical alkanes with even number of carbon atoms.
MCQ 124 Marks
Match the compounds given in list-l with their characteristic reactions in list-ll and choose the correct answer.
| List-l (Compounds) | List-II (Reactions) |
a.  | i. Gives red colouration in victor Meyer's test |
b. | ii. Produces immediate turbidity with Luca's reagent |
| C. CH2CH2CHO | iii. Forms white precipitate with bromine water |
| d . C6H5CH2OH | iv. Forms silver mirror on reaction with ammoniacal AgNO3 solution. |
- Aa(ii), b(iii), c(i), d(iv)
- Ba(iv), b(i), c(ii), d(iii)
- Ca(iii), b(ii), c(iv), d(i)
- ✓a(ii), b(iii), c(iv), d(i)
Answer
View full question & answer→Correct option: D.
a(ii), b(iii), c(iv), d(i)
(D) a(ii), b(iii), c(iv), d(i)
MCQ 134 Marks
Which among the following elements is unable to form a compound in +3 oxidatition state?
- AAl
- ✓ΤI
- CGa
- DIn
Answer
View full question & answer→Correct option: B.
ΤI
(B) ΤI
Tl predominantly exists in +1 oxidation state, due to inert pair effect.
Tl predominantly exists in +1 oxidation state, due to inert pair effect.
MCQ 144 Marks
Assertion(A): Nitrogen exhibits +5 oxidation state
Reason(R): Nitrogen does not form pentahalide
Reason(R): Nitrogen does not form pentahalide
- ABoth (A) and (R) are correct and (R) is the correct explanation of (A)
- ✓Both (A) and (R) are correct but (R) is not the correct explanation of (A)
- C(A) is correct but (R) is not correct
- DBoth the (A) and (R) are not correct.
Answer
View full question & answer→Correct option: B.
Both (A) and (R) are correct but (R) is not the correct explanation of (A)
(B) Both (A) and (R) are correct but (R) is not the correct explanation of (A)
Nitrogen exhibits +5 oxidation state but does not form pentahalides because it does not have dorbitals to expands its covalence beyond four.
Nitrogen exhibits +5 oxidation state but does not form pentahalides because it does not have dorbitals to expands its covalence beyond four.
MCQ 154 Marks
Activation energy of a chemical reaction can be determined by:
- Adetermining the rate constant at standard temperature.
- ✓determining the rate constants at two temperatures.
- Cdetermining probability of collisions.
- Dusing catalyst.
Answer
View full question & answer→Correct option: B.
determining the rate constants at two temperatures.
(B) determining the rate constants at two temperatures.
Arrhenius equation is to be employed.
Arrhenius equation is to be employed.
MCQ 164 Marks
Half-life of a first order reaction with rate constant of $0.02 s^{-1}$ will be nearly :
- A69.3 s
- ✓34.65 s
- C100 s
- D50 s
Answer
View full question & answer→Correct option: B.
34.65 s
(B) 34.65 s
$t_{1 / 2}-\frac{0.693}{k}$
$=\frac{0.693}{0.02}=34.65 s$
$t_{1 / 2}-\frac{0.693}{k}$
$=\frac{0.693}{0.02}=34.65 s$
MCQ 174 Marks
The following mechanism has been proposed for the reaction of NO with $Br _2$ to form NOBr
$\begin{array}{c}
NO(g)+Br_2(g) \rightarrow NOBr_2(g) \\
\left.NOBr_2(g)+NO(g) \rightarrow 2 NOBr^{(g}\right) \end{array}$
If the second step is the rate determining step, the order of reaction w.r.t $NO ( g )$ is:
$\begin{array}{c}
NO(g)+Br_2(g) \rightarrow NOBr_2(g) \\
\left.NOBr_2(g)+NO(g) \rightarrow 2 NOBr^{(g}\right) \end{array}$
If the second step is the rate determining step, the order of reaction w.r.t $NO ( g )$ is:
- A$0$
- B1
- ✓2
- D3
Answer
View full question & answer→Correct option: C.
2
(C) 2
Rate $= k [ NO ]^2\left[ Br _2\right]$
Rate $= k [ NO ]^2\left[ Br _2\right]$
MCQ 184 Marks
Which of the following statements (s) is/are true:
a. Molarity is the number of moles of solute dissolved per litre of solution.
b. Molarity is temperature independent concentration term.
c. Molality of a solution is defined as the number of moles of solute dissolved in 1000 g of solution
d. The ratio of mole fractions of solute and solvent is in the ratio of their respective moles.
a. Molarity is the number of moles of solute dissolved per litre of solution.
b. Molarity is temperature independent concentration term.
c. Molality of a solution is defined as the number of moles of solute dissolved in 1000 g of solution
d. The ratio of mole fractions of solute and solvent is in the ratio of their respective moles.
- Aa and c only
- ✓a and d only
- Cband c only
- Da only
Answer
View full question & answer→Correct option: B.
a and d only
(B) a and d only
MCQ 194 Marks
Which of the following electrolytes will give a linear relationship between molar conductivity $\Lambda_{ m }$ and square root of concentration?
- ACH3COOH
- ✓KCI
- CHCOOH
- DNH4OH
Answer
View full question & answer→Correct option: B.
KCI
(B) KCI
The plot of $\Lambda_\text{mvs } \sqrt{ C }$ for strong electrolyte is a straight line with negative slope.
The plot of $\Lambda_\text{mvs } \sqrt{ C }$ for strong electrolyte is a straight line with negative slope.
MCQ 204 Marks
If $\Lambda_{ m }^{\circ}$ of $NaCl , HCl$ and $CH _3 COONa$ are $x, y$ and $z$ $S cm ^2 mol^{-1}$ respectively then $\Lambda_{ m }^{\circ}$ of $CH _3 COOH$ will be.
- Az + y + x
- Bx - z - y
- ✓z + y - x
- Dz - y - x
Answer
View full question & answer→Correct option: C.
z + y - x
(C) z + y - x
$\begin{aligned} & \wedge_m^o CH _3 COOH =\wedge_m^o CH _3 COO ^{-}+\wedge_m^o H ^{+} \\ = & \wedge_m^o CH _3 COONa +\wedge_m^o HCl -\wedge_m^o NaCl \\ = & z+y-x\end{aligned}$
$\begin{aligned} & \wedge_m^o CH _3 COOH =\wedge_m^o CH _3 COO ^{-}+\wedge_m^o H ^{+} \\ = & \wedge_m^o CH _3 COONa +\wedge_m^o HCl -\wedge_m^o NaCl \\ = & z+y-x\end{aligned}$
MCQ 214 Marks
What is the $E _{\text {cell }}$ of the following adjustment?
$(\log 13=1.119)$
$Ag ( s )\left| Ag ^{+}(0.0001 M ) \| Mg ^{2+}(0.13 M )\right| Mg ( s )$$E _{\text {cell}}^{\circ}=3.17 V$
$Ag ( s )\left| Ag ^{+}(0.0001 M ) \| Mg ^{2+}(0.13 M )\right| Mg ( s )$$E _{\text {cell}}^{\circ}=3.17 V$
- ✓3.38 V
- B2.97 V
- C3.17 V
- D2.75 V
Answer
View full question & answer→Correct option: A.
3.38 V
(A) 3.38 V
$E=E^0=-\frac{0.059}{n} \log Q$
$E=E^0=-\frac{0.059}{n} \log Q$
MCQ 224 Marks
In DNA, C joins with ......... base by ......... hydrogen bonds.
- AA, 2
- ✓G, 3
- CT, 2
- DU, 3
Answer
View full question & answer→Correct option: B.
G, 3
(B) G, 3
In DNA, C joins with G by three hydrogen bonds.
In DNA, C joins with G by three hydrogen bonds.
MCQ 234 Marks
Basic amino acid among the following is:
- ✓Lysine
- BGlutamicacid
- CIsoleucine
- DAlanine
Answer
View full question & answer→Correct option: A.
Lysine
(A) Lysine
Lysine is a basic aminoacid as it has more number
of -NH2 groups than - COOH groups.
Lysine is a basic aminoacid as it has more number
of -NH2 groups than - COOH groups.
Question 244 Marks
$PhNH _2 \xrightarrow{ HNO _2 / 0^{\circ} C } A \xrightarrow{ HBF _4} B \xrightarrow{\Delta} C ; C$ is:
Answer
View full question & answer→(D)
$PhNH _2 \xrightarrow{H N O_2, O_2c} PhN _2 \xrightarrow{\xrightarrow{ } HBE _4} PhN _2^{+}\left( BF _4\right)^{-} \xrightarrow{\Delta} P h-F+ N _2+B F_3$
$PhNH _2 \xrightarrow{H N O_2, O_2c} PhN _2 \xrightarrow{\xrightarrow{ } HBE _4} PhN _2^{+}\left( BF _4\right)^{-} \xrightarrow{\Delta} P h-F+ N _2+B F_3$
MCQ 254 Marks
Sec-butyl alcohol on reaction with I2 and NaOH majorly gives:
- ACH2I2 and CH3 COONa
- ✓CHI3 and CH3 CH2 COONa
- C
- DCl4 and CH3CH2COONa
Answer
View full question & answer→Correct option: B.
CHI3 and CH3 CH2 COONa
(B) CHI3 and CH3 CH2 COONa
MCQ 264 Marks
A solution containing 20 g per dm3 urea is isotonic with a 10% (W/V) solution of a non-volatile solute 'B'. The molar mass of the solute 'B' is.
- A100 g/mol
- B200 g/mol
- ✓300 g/mol
- D600 g/mols
Answer
View full question & answer→Correct option: C.
300 g/mol
(C) 300 g/mol
$\pi=$ CRT
$\pi_1=\pi_2$
$\frac{20}{60 \times 1}=\frac{10 \times 1000}{ Mol. wt _{(B)} \times 100}$
Molar mass of $B=\frac{600}{2}=300 g / mol$
$\pi=$ CRT
$\pi_1=\pi_2$
$\frac{20}{60 \times 1}=\frac{10 \times 1000}{ Mol. wt _{(B)} \times 100}$
Molar mass of $B=\frac{600}{2}=300 g / mol$
MCQ 274 Marks
Out of the following 0.1 molal aqueous solutions which will have maximum freezing point?
- ✓Glucose
- BPotassium chloride
- CSodium sulphate
- DAluminum sulphate
Answer
View full question & answer→Correct option: A.
Glucose
(A) Glucose
$\Delta_f T=i m K_f$
For the solutions of same $m$ and $K_f$
$\Delta_f T \sim i$
For $C _6 H _{12} O _6 \quad \Rightarrow i =1$
$KCl \Rightarrow i =2$
$Na _2 SO _4 \Rightarrow i =3$
$Al _2 SO _4 \Rightarrow i =5$
Minimum $i$ is for glucose so $\Delta T_f$ will be minimum for it and freezing point will be maximum for glucose.
$\Delta_f T=i m K_f$
For the solutions of same $m$ and $K_f$
$\Delta_f T \sim i$
For $C _6 H _{12} O _6 \quad \Rightarrow i =1$
$KCl \Rightarrow i =2$
$Na _2 SO _4 \Rightarrow i =3$
$Al _2 SO _4 \Rightarrow i =5$
Minimum $i$ is for glucose so $\Delta T_f$ will be minimum for it and freezing point will be maximum for glucose.
Question 284 Marks
Correct order of boiling points for the given compounds is
Answer
View full question & answer→(C) 3.38 V
MCQ 294 Marks
The CFSE for octahedral $\left[ CoCl _6\right]^{+}$is corresponding to 18,000 (w.r.t wave number $cm ^{-1}$ ). The CFSE for tetrahedral $\left[ CoCl _4\right]^{2-}$ will be:
- A18000 cm -1
- B16000 cm -1
- ✓8000 cm -1
- D20000 cm -1
Answer
View full question & answer→Correct option: C.
8000 cm -1
(C) 8000 cm -1
$\Delta \propto \bar{v}$
$\bar{v}_0=\frac{4}{9} \bar{v}_t$
$\Delta \propto \bar{v}$
$\bar{v}_0=\frac{4}{9} \bar{v}_t$
MCQ 304 Marks
Match list-l with list-ll and choose the correct option.
| List-l(Compound) | List-ll (isomerism) |
| (a) $\left[ Co \left( NH _3\right)_6\right]\left[ Cr ( CN )_6\right]$ | i. Optical isomerism |
| (b) $\left[ Fe \left( H _2 O \right)_5 Cl \right] Br$ | ii. Coordination isomerism |
| (c) $\left[ PtCl _2\left( NH _3\right)_2\right]$ | iii. lonisation isomerism |
| (d) $K _3\left[ Co ( OX )_3\right]$ | iv. Geometrical isomerism |
- Aa(ii), b(iii), c(i), d(iv)
- ✓a(ii), b(iii), c(iv), d(i)
- Ca(iii), b(ii), c(i), d(iv)
- Da(iii), b(ii), c(iv), d(i)
Answer
View full question & answer→Correct option: B.
a(ii), b(iii), c(iv), d(i)
(B) a(ii), b(iii), c(iv), d(i)
- $\left[ Co \left( NH _3\right)_{6}\right]\left[ Cr ( CN )_6\right]$ can show Coordination isomerism
- [Fe(H2O)5Cl]Br can show Ionisation isomerism
- $\left[ PtCl _2\left( NH _3\right)_2\right]$ can show Geometrical isomerism
- $K _3\left[ Co ( OX )_3\right]$ can show Optical isomerism
- $\left[ Co \left( NH _3\right)_{6}\right]\left[ Cr ( CN )_6\right]$ can show Coordination isomerism
- [Fe(H2O)5Cl]Br can show Ionisation isomerism
- $\left[ PtCl _2\left( NH _3\right)_2\right]$ can show Geometrical isomerism
- $K _3\left[ Co ( OX )_3\right]$ can show Optical isomerism
MCQ 314 Marks
Ambidentate ligand among the following is:
- AH2O
- Ben
- CNH3
- ✓CN-
Answer
View full question & answer→Correct option: D.
CN-
(D) CN-
Both carbon and nitrogen can work as donating site in CN hence it is an ambidentate ligand.
Both carbon and nitrogen can work as donating site in CN hence it is an ambidentate ligand.
MCQ 324 Marks
Correct order of ionic radii of the given species is
- ✓$Nd ^{3+}> Tb ^{3+}> Tm ^{3+}$
- B$Tm ^{3+}> Tb ^{3+}> Nd ^{3+}$
- C$Tb ^{3+}> Nd ^{3+}> Tm ^{2+}$
- D$Tb ^{3+}> Tm ^{3+}> Nd ^{3+}$
Answer
View full question & answer→Correct option: A.
$Nd ^{3+}> Tb ^{3+}> Tm ^{3+}$
(A)$Nd ^{3+}> Tb ^{3+}> Tm ^{3+}$
MCQ 334 Marks
Assertion(A): Pbl4 is a stable compound.
Reason(R): lodide stabilises higher oxidation state.
Reason(R): lodide stabilises higher oxidation state.
- ABoth (A) and (R) are correct and (R) is the correct explanation of (A)
- BBoth (A) and (R) are correct but (R) is not the correct explanation of (A)
- C(A) is correct but (R) is not correct
- ✓Both the (A) and (R) are not correct.
Answer
View full question & answer→Correct option: D.
Both the (A) and (R) are not correct.
(D) Both the (A) and (R) are not correct.
$Pbi _4$ is not stable.
$Pbi _4$ is not stable.
MCQ 344 Marks
Given below are two statements.
Statement-l: Negative electron gain enthalpy of fluorine is greater than chlorine.
Statement-II: lonisation enthalpy of fluorine is less than chlorine.
In the light of above statements, choose the most appropriate answer from the options given below.
Statement-l: Negative electron gain enthalpy of fluorine is greater than chlorine.
Statement-II: lonisation enthalpy of fluorine is less than chlorine.
In the light of above statements, choose the most appropriate answer from the options given below.
- ABoth statement I and statement II are correct
- ✓Both statement I and statement II are incorrect
- CStatement I is correct but statement II is incorrect
- DStatement I is incorrect but statement II is correct.
Answer
View full question & answer→Correct option: B.
Both statement I and statement II are incorrect
(B) Both statement I and statement II are incorrect
Negative electron gain enthalpy of fluorine is less than chlorine while is more than the chlorine.
Negative electron gain enthalpy of fluorine is less than chlorine while is more than the chlorine.
MCQ 354 Marks
Maximum number of electrons in a subshell having I = 2 and n = 3 is
- A2
- B6
- ✓10
- D14
Answer
View full question & answer→Correct option: C.
10
(C) 10
Subshell = 3d
Maximum number of electrons in d subshell is 10.
Subshell = 3d
Maximum number of electrons in d subshell is 10.
MCQ 364 Marks
If the mass of an electron is 9.11x10-31 kg, and uncertainty in position is 10-8 m, the uncertainty in velocity would be:
- A5.8 x10-3 ms-1
- ✓5.8 x103 ms-1
- C5.8 x10-8 ms-1
- D5.8 x108 ms-1
Answer
View full question & answer→Correct option: B.
5.8 x103 ms-1
(B) 5.8 x103 ms-1
$\Delta v=\frac{h}{4 \pi m \Delta x}=\frac{6.626 \times 10^{-34} Js }{4 \times \frac{22}{7} \times 9.11 \times 10^{-31} kg \times 10^{-8} m}$
$\Delta v=\frac{h}{4 \pi m \Delta x}=\frac{6.626 \times 10^{-34} Js }{4 \times \frac{22}{7} \times 9.11 \times 10^{-31} kg \times 10^{-8} m}$
MCQ 374 Marks
For any H like species, the ratio of radii of $1 ^{\text {st }}, 2 ^{\text {nd }}$ and $3^{\text {rd }}$ orbits is
- ✓(1:4:9)
- B$1: \frac{1}{4}: \frac{1}{9}$
- C(1:2:3)
- D(1:1:1)
Answer
View full question & answer→Correct option: A.
(1:4:9)
(A) (1:4:9)
$r_n \propto n^2$
$r_n \propto n^2$
Question 384 Marks
Answer
View full question & answer→(A)
MCQ 394 Marks
Match list-l with list-ll and choose the correct answer.
| List -I (Group attached in benzene) | List-ll (Effect shown by the group) |
| a. -NO2 | (i) -R effect and -l effect |
| b. -O- | (ii) +R effect and -l effect |
| c. -O-CH3 | (iii)+R effect and +l effect |
| d. -CH3 | (iv) +1 effect and no resonance |
- ✓a(i) b(ii) c(ii) d(iv)
- Ba(i) b(ii) c(iii), d(iv)
- Ca(Iv) b(iii) c(ii) d(i)
- Da(i) b(iv) c(iii), d(ii)
Answer
View full question & answer→Correct option: A.
a(i) b(ii) c(ii) d(iv)
(A) a(i) b(ii) c(ii) d(iv)
$- NO _2 \rightarrow- R$ and - l effect
$- O ^{-} \rightarrow+ R$ and +1 effect
$- O - CH _3 \rightarrow+ R$ and -1 effect
$- CH _3 \rightarrow+1$ effect.
$- NO _2 \rightarrow- R$ and - l effect
$- O ^{-} \rightarrow+ R$ and +1 effect
$- O - CH _3 \rightarrow+ R$ and -1 effect
$- CH _3 \rightarrow+1$ effect.
MCQ 404 Marks
For the equillibrium, $C ( s )+ CO _2(g) \rightleftharpoons 2 CO ( g )$ $k_p=60 atm$ at 1000 K . If at equillibrium $P _{ co }=10 P _{ co 2}$, then the total pressure at equillibrium is
- A6.1 atm
- ✓6.6 atm
- C0.6 atm
- D66.6 atm
Answer
View full question & answer→Correct option: B.
6.6 atm
(B) 6.6 atm
$C ( s )+ CO _2(g) \rightleftharpoons 2 CO ( g )$
$\frac{\left(10 P _{ CO _2}\right)^2}{ P _{ CO _2}}=60$
$C ( s )+ CO _2(g) \rightleftharpoons 2 CO ( g )$
$\frac{\left(10 P _{ CO _2}\right)^2}{ P _{ CO _2}}=60$
MCQ 414 Marks
In a compound, C-40% and H-6.66% are present, what is the empirical formula of the compound:
- ✓CH2O
- BC4HO4
- CCHO
- DCH2O2
Answer
View full question & answer→Correct option: A.
CH2O
(A) CH2O
MCQ 424 Marks
The bond energies for $A - A , B - B$ and $A - B$ are $a , b, c$ and kj respectively. The $\Delta_{ r } H$ for the reaction $\frac{1}{2} A_2+\frac{1}{2} B_2 \rightarrow A B$ will be :
- Aa + b - c
- B$c-\frac{a}{2}-\frac{b}{2}$
- Cc - a - b
- ✓$\frac{a}{2}+\frac{b}{2}-c$
Answer
View full question & answer→Correct option: D.
$\frac{a}{2}+\frac{b}{2}-c$
(D) $\frac{a}{2}+\frac{b}{2}-c$
$\frac{1}{2} A_2+\frac{1}{2} B_2 \rightarrow A B$
$\Delta_r H=\sum B E_{\text {Reac tant }}-\sum B E_{\text {products }}$
$\frac{1}{2} a+\frac{1}{2} b-c$
$\Delta_r H=\frac{a}{2}+\frac{b}{2}-c$
$\frac{1}{2} A_2+\frac{1}{2} B_2 \rightarrow A B$
$\Delta_r H=\sum B E_{\text {Reac tant }}-\sum B E_{\text {products }}$
$\frac{1}{2} a+\frac{1}{2} b-c$
$\Delta_r H=\frac{a}{2}+\frac{b}{2}-c$
MCQ 434 Marks
Using molecular orbital theory, predict which of the following species has the shortest bond length?
- ✓$O _2^{+}$
- B$O _2^{-}$
- C$O _2{ }^{2-}$
- D$O _2$
Answer
View full question & answer→Correct option: A.
$O _2^{+}$
(A) $O _2^{+}$
Bond length decreases as bond order increases.
| Species | Bond Order |
| (a) $O _2^{+}$ | (i) 2.5 |
| (b) $O _2^{-}$ | (ii) 1.5 |
| (c) $O _2^{2+}$ | (iii) 1 |
| (d) $O_2$ | (iv) 2 |
MCQ 444 Marks
Which of the following molecules has the maximum dipole moment ?
- ✓NH3
- BNF3
- CCO2
- DCH4
Answer
View full question & answer→Correct option: A.
NH3
(A) NH3
MCQ 454 Marks
Which branched chain isomer of hydrocarbon with molecular mass 72 u gives only one mono substituted alkyl halide
- AIsopentane
- ✓Neopentane
- Chexane
- DNeohexene
Answer
View full question & answer→Correct option: B.
Neopentane
(B) Neopentane
