Chemistry Test 156

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Chemistry Test 156

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Weekly Quiz Competition

Question 1
4 / -1

O_(g) + 2e^– → O^2–_(g) ΔHeg = 603 KJ/mole. The positive value of ΔHeg is due to :-

A
Energy is released to add on 1 e^– to O^–1

B
Energy is required to add on 1 e^– to O^–1

C
Energy is needed to add on 1e^– to O

D
None of the above is correct
Question 2
4 / -1

Identify the incorrect statement related to PCl₅ from the following :-

A
Three equatorial P–Cl bonds make an angle of 120° with each other

B
Two axial P–Cl bonds make an angle of 180° with each other

C
Axial P–Cl bonds are longer than equatorial P–Cl bonds

D
PCl₅ molecule is non–reactive
Question 3
4 / -1

Which of the following diatomic molecular species has only π bonds according to Molecular Orbital Theory ?

A
O₂

B
N₂

C
C₂

D
Be₂
Question 4
4 / -1

Berylium hydroxide dissolves in excess of alkali to give

A

B

C

D
does not react with alkali because of its basic nature
Question 5
4 / -1

Incorrect reaction is-

A
XeF₆ + 3 H₂O → XeO₃ + HF

B
XeF₆ + H₂O → XeOF₄ + HF

C
XeF₆ + 2 H₂O → XeO₂F₂+ HF

D
XeF₆ + 2 H₂O → Xe + HF + O₂
Question 6
4 / -1

True statement is

A
X can't be MnO₂

B
Y can be KMnO₄

C
Z can be K₂MnO₄

D
X is MnO₂ and Y is K₂MnO₄
Question 7
4 / -1

An incorrect statement is :

A
The metal-carbon bond in metal carbonyls posess both s and p character

B
The M-C π bond is formed by the donation of a pair of electrons from a filled d orbital of metal into the vacant antibonding π* orbital of carbon monoxide

C
The metal to ligand bonding creates a synergic effect which strengthens the bond between C – O in CO

D
Due to M-C π bond the C – O bond order decreases and M – C bond length decreases
Question 8
4 / -1

A crystal is made of particles X, Y and Z. X forms fcc packing. Y occupies all the octahedral voids of X and Z occupies all the tetrahedral voids. If all the particles along one body diagonal are removed then the formula of the crystal would be :-

A
XYZ

B
X₂YZ₂

C
X₈Y₄Z₅

D
X₅Y₄Z₈

Solution

X → fcc = 6 × 1/8 + 6 × 1/2 = 15/4

Y → OHV = 12 × 1/4 = 3

Z → THV = 6

X_15/4 Y₃ Z₆

X₁₅ Y₁₂ Z₂₄ ⇒ X₅ Y₄ Z₈
Question 9
4 / -1

Directions For Questions

The coagulation value in millimoles per litre of electrolytes used for the coagulation of As₂S₃ are as below :-

I. NaCl = 52 II. KCl = 51

III. BaCl₂ = 0.69 IV. MgSO₄ = 0.22

...view full instructions

The correct order of their flocculating power is :

A
I > II > III > IV

B
I > II > III = IV

C
IV > III > II > I

D
IV = III > II > I

Solution

↑ will be coagulation power ↓ cogulation value
Question 10
4 / -1

An ideal mixture of liquids A and B with 2 moles of A and 2 moles of B has a total vapour pressure of 1 atm at a certain temperature. Another mixture with 1 mole of A and 3 moles of B has a vapour pressure greater than 1 atm. But if 4 moles of C are added to the second mixture, the vapour pressure comes down to 1 atm. Vapour pressure of C, P_c⁰ = 0.8 atm. Calculate the vapour pressures of pure A and pure B :-

A
P_A⁰ = 1.4 atm, P_B⁰ = 0.7 atm

B
P_A⁰ = 1.2 atm, P_B⁰ = 0.6 atm

C
P_A⁰ = 1.4 atm, P_B⁰ = 0.6 atm

D
P_A⁰ = 0.6 atm, P_B⁰ = 1.4 atm

Solution

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Chemistry Test 156

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Question 1

Energy is released to add on 1 e– to O–1

Energy is required to add on 1 e– to O–1

Energy is needed to add on 1e– to O

None of the above is correct

Question 2

Three equatorial P–Cl bonds make an angle of 120° with each other

Two axial P–Cl bonds make an angle of 180° with each other

Axial P–Cl bonds are longer than equatorial P–Cl bonds

PCl5 molecule is non–reactive

Question 3

O2

N2

C2

Be2

Question 4

does not react with alkali because of its basic nature

Question 5

XeF6 + 3 H2O → XeO3 + HF

XeF6 + H2O → XeOF4 + HF

XeF6 + 2 H2O → XeO2F2+ HF

XeF6 + 2 H2O → Xe + HF + O2

Question 6

X can't be MnO2

Y can be KMnO4

Z can be K2MnO4

X is MnO2 and Y is K2MnO4

Question 7

The metal-carbon bond in metal carbonyls posess both s and p character

The M-C π bond is formed by the donation of a pair of electrons from a filled d orbital of metal into the vacant antibonding π* orbital of carbon monoxide

The metal to ligand bonding creates a synergic effect which strengthens the bond between C – O in CO

Due to M-C π bond the C – O bond order decreases and M – C bond length decreases

Question 8

XYZ

X2YZ2

X8Y4Z5

X5Y4Z8

Solution

Question 9

I > II > III > IV

I > II > III = IV

IV > III > II > I

IV = III > II > I

Solution

Question 10

PA0 = 1.4 atm, PB0 = 0.7 atm

PA0 = 1.2 atm, PB0 = 0.6 atm

PA0 = 1.4 atm, PB0 = 0.6 atm

PA0 = 0.6 atm, PB0 = 1.4 atm

Solution

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Energy is released to add on 1 e^– to O^–1

Energy is required to add on 1 e^– to O^–1

Energy is needed to add on 1e^– to O

PCl₅ molecule is non–reactive

O₂

N₂

C₂

Be₂

XeF₆ + 3 H₂O → XeO₃ + HF

XeF₆ + H₂O → XeOF₄ + HF

XeF₆ + 2 H₂O → XeO₂F₂+ HF

XeF₆ + 2 H₂O → Xe + HF + O₂

X can't be MnO₂

Y can be KMnO₄

Z can be K₂MnO₄

X is MnO₂ and Y is K₂MnO₄

X₂YZ₂

X₈Y₄Z₅

X₅Y₄Z₈

P_A⁰ = 1.4 atm, P_B⁰ = 0.7 atm

P_A⁰ = 1.2 atm, P_B⁰ = 0.6 atm

P_A⁰ = 1.4 atm, P_B⁰ = 0.6 atm

P_A⁰ = 0.6 atm, P_B⁰ = 1.4 atm