Mix Test

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Mix Test - 2

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following has highest ionisation enthalpy?

A
Nitrogen

B
Phosphorus

C
Oxygen

D
Sulphur

Solution

Nitrogen (High IE of N is because of smallest size in the group and completely half - filled p subshell).
Question 2
1 / -0

Metal M ions form accp structure. Oxide ions occupy \(\frac12\) octahedral and \(\frac12\) tetrahedral voids. What is the formula of the oxide?

A
MO

B
M\(O_2\)

C
M\(O_3\)

D
\(M_2O_3\)

Solution

\(M_2O_3\)

Metal M ions form ccp structure. Let number of ions of M be : X

No. of tetrahedral voids = 2x

No. of octahedral voids = x

Number of oxide ions will be \(\frac12\) x + \(\frac12\) (2x) = 3/2 x

Formula of oxide = \(\frac{M\text xO_3}{2 \text x}\) = \(M_2O_3\)
Question 3
1 / -0

The reaction of toluene with \(Cl_2\) in presence of Fe\(Cl_3\) gives ‘X’ while theof toluene with \(Cl_2\) in presence of light gives ‘Y’. Thus ‘X’ and ‘Y’are:

A
X = benzyl chloride Y = o and p – chlorotoluene

B
X = m – chlorotoluene Y = p – chlorotoluene

C
X = o and p–chlorotoluene Y = trichloromethylbenzene

D
X= benzyl chloride, Y = m-chlorotoluene

Solution

X = o and p–chlorotoluene Y = trichloromethylbenzene

The reaction of toluene with \(Cl_2\) in presence of Fe\(Cl_3\) gives ‘X’ due to electrophilic substitution reaction taking place at ortho and para positions and reaction in the presence of light gives ‘Y’, due to substitution reaction occurring via free radical mechanism .

Thus ‘X’ and ‘Y’are X = o and p–chlorotoluene Y = trichloromethylbenzene.
Question 4
1 / -0

Ozone is a/ an __________ molecule and the two O-O bond lengths in ozone are (i)_______-and (ii) ____________.

A
linear ,110pm ; 148pm

B
angular, 110pm ; 148pm

C
linear, 128pm ; 128pm

D
angular, 128pm ; 128pm

Solution

angular, 128pm ; 128pm (Ozone is a resonance hybrid of two equivalent structures).
Question 5
1 / -0

Water retention or puffiness due to high salt intake occurs due to:

A
diffusion

B
vapour pressure difference

C
osmosis

D
reverse osmosis

Solution

Osmosis.
Question 6
1 / -0

In lake test for \(Al^{3+}\) ions, there is the formation of coloured ‘floating lake’. It is due to:

A
Absorption of litmus by\( [Al(OH)_4]^ -\)

B
Absorption of litmus by \(Al(OH)_3\)

C
Adsorption of litmus by \([Al(OH)4]^-\)

D
Adsorption of litmus by \(Al(OH)_3\)

Solution

Adsorption of litmus by \(Al(OH)_3\)

In lake test for \(Al^{3+}\) ions, there is the formation of coloured ‘floating lake’ In lake test for \(Al^{3+}\) ions, there is the formation of coloured ‘floating lake’ due to adsorption.
Question 7
1 / -0

A unit cell of NaCl has 4 formula units. Its edge length is 0.50 nm. Calculate the density if molar mass of NaCl = 58.5 g/mol.

A
1 g/\(cm^3\)

B
2 g/\(cm^3\)

C
3 g/\(cm^3\)

D
4g/\(cm^3\)

Solution

3 g/\(cm^3\)

Using formula

Density = \(\frac{(Z\,\times\,m)}{(a^3\,\times\,Na)}\)

D = \(\frac{4\times58.5}{(0.5\times10^{-7})^3\times6.023\times10^{23}}\)

= 3.1 g/\(cm^3\)
Question 8
1 / -0

Which one of the following are correctly arranged on the basis of the property indicated:

A
\(I_2<Br_2<F_2<Cl_2\) [increasing bond dissociation enthalpy]

B
\(H_2O > H_2S<H_2Te<H_2Se\) [increasing acidic strength]

C
\(NH_3 < N_2O< NH_2OH<N_2O_5\) [ increasing oxidation state]

D
\(BiH_3<SbH_3<AsH_3<PH_3<NH_3\) [ increasing bondangle]

Solution

\(BiH_3<SbH_3<AsH_3<PH_3<NH_3\) [increasing bond angle due to leaving ammonia other are drago molecules which donot show hybridization and thus from bond angle close to 90°] - correct order

(a) \(I_2<Br_2<F_2<Cl_2\) [increasing bond dissociation enthalpy]: incorrect order , correct order is \(Cl_2 > Br_2 > F_2 > I_2.\)

(b) \(H_2O > H_2S<H_2Te<H_2Se\) [increasing acidic strength]: incorrect order , correct order is \(H_2O<H_2S<H_2Se<H_2Te\)

(c) \(NH_3 < N_2O< NH_2OH<N_2O_5\) [ increasing oxidation state] : incorrect order \(NH_3\) (Oxidation state -3) \(N_2O\) (Oxidation state +1) \(NH_2OH\)(Oxidation state -1) \(N_2O_5\) (Oxidation state +5).
Question 9
1 / -0

Which of the following statements is true:

A
Ammonia is the weakest reducing agent and the strongest base among Group 15 hydrides.

B
Ammonia is the strongest reducing agent as well as the strongest base among Group 15 hydrides.

C
Ammonia is the weakest reducing agent as well as the weakest base among Group 15 hydrides.

D
Ammonia is the strongest reducing agent and the weakest base among Group 15 hydrides.

Solution

Ammonia is the weakest reducing agent and the strongest base among Group 15 hydrides. The reducing character of hydrides increases down the group due to decrease in bond dissociation enthalpy.
Question 10
1 / -0

Alkenes decolourise bromine water in presence of \(CCl_4\) due to formation of:

A
allyl bromide

B
vinyl bromide

C
bromoform

D
vicinal dibromide

Solution

vicinal dibromide

\(CH_2=CH_2 + Br_2 \longrightarrow BrCH_2 - CH_2Br\)
Question 11
1 / -0

Assertion (A): Electron gain enthlpy of oxygen is less than that of Flourine but greater than Nitrogen.

Reason (R): Ionisation enthalpies of the elements follow the order Nitrogen > Oxygen > Fluorine.

A
Both A and R are true and R is the correct explanation of A

B
Both A and R are true but R is not the correct explanation of A.

C
A is true but R is false.

D
A is true but R is false.

Solution

Assertion: Electron gain enthalpy of oxygen is less than that of Flourine but greater than Nitrogen. (correct)

Reason: Ionisation enthalpies of the elements follow the order Nitrogen > Oxygen > Fluorine (incorrect)

Ionisation enthalpies of the elements follow the order Fluorine >Nitrogen > Oxygen
Question 12
1 / -0

Assertion (A): Alkyl halides are insoluble in water.

Reason (R): Alkyl halides have halogen attached to \(sp^3\) hybrid carbon. Select the most appropriate answer from the options given below:

A
Both A and R are true and R is the correct explanation of A

B
Both A and R are true but R is not the correct explanation of A.

C
A is true but R is false.

D
A is false but R is true.

Solution

Assertion: Alkyl halides are insoluble in water. (correct)

Reason: Alkyl halides have halogen attached to \(sp^3\) hybrid carbon. (correct) Alkyl halides are insoluble in water because they are unable to form hydrogen bonds with water or break pre-existing hydrogen bonds.
Question 13
1 / -0

Assertion(A): Molarity of a solution changes with temperature.

Reason (R): Molarity is a colligative property.

A
Both A and R are true and R is the correct explanation of A

B
Both A and R are true but R is not the correct explanation of A.

C
A is true but R is false.

D
A is false but R is true.

Solution

Assertion: Molarity of a solution changes with temperature. (correct)

Reason: Molarity is a colligative property. (incorrect)

Molarity is a means to express concentration. It is not a physical property.
Question 14
1 / -0

Assertion(A): \(SO_2\) is reducing while \(TeO_2\) is an oxidising agent.

Reason(R): Reducing property of dioxide decreases from \(SO_2\) to \(TeO_2\).

A
Both A and R are true and R is the correct explanation of A

B
Both A and R are true but R is not the correct explanation of A.

C
A is true but R is false

D
A is false but R is true.

Solution

Assertion: \(SO_2\) is reducing while \(TeO_2\) is an oxidising agent. (correct)

Reason: Reducing property of dioxide decreases from \(SO_2\) to \(TeO_2\) (correct and reason for Assertion)
Question 15
1 / -0

Assertion (A):Cryoscopic constant depends on nature of solvent.

Reason(R ):Cryoscopic constant is a universal constant.

A
Both A and R are true and R is the correct explanation of A

B
Both A and R are true but R is not the correct explanation of A.

C
A is true but R is false.

D
A is false but R is true.

Solution

Assertion: Cryoscopic constant depends on nature of solvent. (correct)

Reason: Cryoscopic constant is a universal constant (incorrect)

Cryoscopic constant various with type of solvent.

Question 16

1 / -0

Match the following:

I	II
(i)Amino acids	(A)protein
(ii)Thymine	(B)Nucleic acid
(iii)Insulin	(C)DNA
(iv)phosphodiester linkage	(D)Zwitter ion
(v) Uracil

Which of the following is the best matched options?

i-A, v- D, iii- C, iv-B

i-D, ii-C, iii- A, iv-B

i-D, v- D, iii- A, iv-B

i-A, ii- C, iii- D, iv-B

Solution

i-D, ii-C, iii- A, iv-B

Amino acids form proteins and exist as zwitter ion , Thymine is a nitrogenous base in DNA, Insulin is a protein , phosphodiester linkage is found in nucleic acids so also in DNA and Uracil is nitrogenous base found in RNA which is a nucleic acid.

Question 17
1 / -0

Which of the following analogies is correct:

A
Nitrogen: \(1s^2 2s^2 2p^3 :: Argon:1s^2 2s^2 2p^6\)

B
Carbon: maximum compounds :: Xenon: no compounds

C
\(XeF_2\) : Linear :: \(ClF_3\) : Trigonal planar

D
Helium: meteorological observations:: Argon: metallurgical processes

Solution

Helium: meteorological observations :: Argon: metallurgical processes

Nitrogen: \(1s^2 2s^2 2p^3 :: Argon:1s^2 2s^2 2p^6\) is configuration of Neon not Argon

Carbon: maximum compounds :: Xenon: no compounds , Xenon forms compounds

\(XeF_2\) : Linear :: \(ClF_3\) : Trigonal planar , \(ClF_3\) is T shaped not trigonal planar.
Question 18
1 / -0

Complete the following analogy:

Same molecular formula but different structures: A:: Non superimposable mirror images: B

A
A:Isomers B: Enantiomer

B
A: Enantiomers B: Racemic mixture

C
A: Sterioisomers B: Retention

D
A: Isomers B: Sterioisomers

Solution

A : Isomers B: Enantiomer

Isomers have Same molecular formula but different structure

Enantiomers are Non superimposable mirror images.
Question 19
1 / -0

The radius of \(Ag^+\) ion is 126 pm and of \(I^-\) ion is 216pm. The coordination number of \(Ag^+\) ion is:

A
2

B
3

C
6

D
8

Solution

The radius of \(Ag^+\) ion is 126pm and of \(I^-\)ion is 216pm. The coordination number of \(Ag^+\) ion is:

\(\rho=\frac{r_{cation}}{r_{anion}}\) = \(\frac{126}{126}\) = 0.58

Radius ratio lies in the range 0.414 – 0.732, so has coordination number 6 or 4 according to the table.

Since none of the options is 4, so the answer is 6
Question 20
1 / -0

A solid AB has square planar structure. If the radius of cation is 120 pm, calculate the maximum possible value of anion

A
240 pm

B
270 pm

C
280 pm

D
290 pm

Solution

290 pm

Square planar means ratio ratio is between 0.414 – 0.732

If radius of cation is 120 pm then anion should be in the range

\(\rho= \frac{r_{cation}}{r_{anion}}\)

0.414 = \(\frac{120}{\text x}\) so x = 289.8 = 290 pm

0.732 = \(\frac{120}{\text x}\) so x = 163.9 = 164 pm
Question 21
1 / -0

A “good fit” is considered to be one where the cation can touch:

A
all of its nearest neighbour anions.

B
most of its nearest neighbour anions.

C
some of its nearest neighbour anions.

D
none of its nearest neighbour anions.

Solution

All of its nearest neighbour anions.

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Mix Test - 2

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Mix Test - 2

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Question 1

Nitrogen

Phosphorus

Oxygen

Sulphur

Solution

Question 2

MO

M\(O_2\)

M\(O_3\)

\(M_2O_3\)

Solution

Question 3

X = benzyl chloride Y = o and p – chlorotoluene

X = m – chlorotoluene Y = p – chlorotoluene

X = o and p–chlorotoluene Y = trichloromethylbenzene

X= benzyl chloride, Y = m-chlorotoluene

Solution

Question 4

linear ,110pm ; 148pm

angular, 110pm ; 148pm

linear, 128pm ; 128pm

angular, 128pm ; 128pm

Solution

Question 5

diffusion

vapour pressure difference

osmosis

reverse osmosis

Solution

Question 6

Absorption of litmus by\( [Al(OH)_4]^ -\)

Absorption of litmus by \(Al(OH)_3\)

Adsorption of litmus by \([Al(OH)4]^-\)

Adsorption of litmus by \(Al(OH)_3\)

Solution

Question 7

1 g/\(cm^3\)

2 g/\(cm^3\)

3 g/\(cm^3\)

4g/\(cm^3\)

Solution

Question 8

\(I_2<Br_2<F_2<Cl_2\) [increasing bond dissociation enthalpy]

\(H_2O > H_2S<H_2Te<H_2Se\) [increasing acidic strength]

\(NH_3 < N_2O< NH_2OH<N_2O_5\) [ increasing oxidation state]

\(BiH_3<SbH_3<AsH_3<PH_3<NH_3\) [ increasing bondangle]

Solution

Question 9

Ammonia is the weakest reducing agent and the strongest base among Group 15 hydrides.

Ammonia is the strongest reducing agent as well as the strongest base among Group 15 hydrides.

Ammonia is the weakest reducing agent as well as the weakest base among Group 15 hydrides.

Ammonia is the strongest reducing agent and the weakest base among Group 15 hydrides.

Solution

Question 10

allyl bromide

vinyl bromide

bromoform

vicinal dibromide

Solution

Question 11

Both A and R are true and R is the correct explanation of A

Both A and R are true but R is not the correct explanation of A.

A is true but R is false.

A is true but R is false.

Solution

Question 12

Both A and R are true and R is the correct explanation of A

Both A and R are true but R is not the correct explanation of A.

A is true but R is false.

A is false but R is true.

Solution