Chemistry Test 137

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Chemistry Test 137

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Weekly Quiz Competition

Question 1
4 / -1

If the nitrogen atom had electronic configuration 1s⁷, it would have energy lower than that of the normal ground state configuration 1s² 2s²2p³, because the electrons would be closer to the nucleus, yet 1s⁷ is not observed because it violates

A
Heisenberg uncertainty principle

B
Aufbau rule

C
Pauli exclusion principle

D
Bohr postulate of stationary orbits

Solution

1s⁷violate Pauli exclusion principle, according to which an orbital cannot have more than two electrons.
Question 2
4 / -1

Amongst the following, the most basic compound is

A
Aniline

B
Acetanilide

C
p-nitroaniline

D
Benzylamine

Solution

Lone pair is not involved in resonance, most basic. In all other cases, lone-pair of nitrogen is involved in resonance, less basic.
Question 3
4 / -1

How many EDTA (ethylenediaminetetraacetic acid) molecules are required to make an octahedral complex with a Ca²⁺

A
Six

B
Three

C
One

D
Two

Solution

EDTA, Which has four lone pair donor oxygen atoms and two lone pair donor nitrogen atoms in each molecule forms complex with Ca²⁺ ion. EDTA is hexa due to need 1 EDTA to form octahedral completes with ca⁺²
Question 4
4 / -1

The first orbital of H is represented by

where a₀ is Bohr's radius The probability of finding the electron at a distance r, from the nucleus in the region dV is

A
∫ψ²4πr²dv

B
∫ψdv

C
ψ²dr

D
ψ²4πr²dr

Solution

P(r) = ψ²4πr²dr
4πr²dr = dv = vol of thin spherical shell around nucleus at distance (r)
Question 5
4 / -1

20% of N₂O₄ molecules are dissociated in a sample of gas at 27^∘C and 760 torr. Mixture has the density at equilibrium equal to:

A
1.48 g/L

B
1.84 g/L

C
2.25 g/L

D
3.12 g/L

Solution

Total Moles 1 + α at equilibrium observed Molar Mass at equilibrium
Question 6
4 / -1

Which of the following compounds display geometrical isomerism?

A
H₂C = CHBr

B
H₂C = CBr₂

C
(Cl)HC = CH(Br)

D
Br₂C = CCl₂

Solution

Apart from this option, the other compounds don’t show geometrical isomerism.
Question 7
4 / -1

Nickel (Z = 28) combines with a uninegative mono dentate ligand X- to form a paramagnetic complex [NiX₄]^2-.The number of unpaired electron(s) in the nickel and geometry of this complex ion are, respectively

A
One, tetrahedral

B
One, square planar

C
Two, tetrahedral

D
Two, square planar

Solution

Number of unpaired electrons = 2
Geometry = tetrahedral.
Question 8
4 / -1

The correct stability order of the following resonance structures is?

A
(i) > (ii) > (iv) > (iii)

B
(i) > (ii) > (iii) > (iv)

C
(iii) > (i) > (iv) > (ii)

D
(i) > (iii) > (ii) > (iv)

Solution

No. of π π bonds ∝∝ Resonance Energy ∝∝ Stability

+ve charge on electro positive element

-ve charge on electro negative element

unlike charges on adjacent atoms like charges to separate out far.

(i) Two π π bonds unlike charges on adjacent atoms

(iii) Two π π Bonds unlike charges on adjacent undesired atoms

(ii) One π π bond unlike charges separated but on desired atoms.

(iv) One π π bond unlike charges separated but on undesired atoms

(i) > (iii) > (ii) > (iv)
Question 9
4 / -1

Identify Z in the sequence,

A

B

C
CH₃- (CH₂)₃ - O - CH₂ - CH

D
CH₃ - (CH₂)₄ - O - CH₃

Solution

In presence of H₂O₂, HBr adds in anti-Markownikoffs way (peroxide effect).
Question 10
4 / -1

The shape of XeF₄ is :

A
Tetrahedral

B
Square planar

C
Octahedral

D
Trigonal planar

Solution

Due to presence of two lone pairs of electrons, it is square planar in shape. Lone pairs are arranged at pyramidal position as per Bent's Rule.