Chemistry Test 129

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Chemistry Test 129

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Weekly Quiz Competition

Question 1
4 / -1

Which of the following can act as reducing agent

A
[Co(CO)₄]^–

B
[Mn(CO)₆]

C
[Mn(CO)₅]

D
[Cr(CO)₆]

Solution

Stability of metal carbonyl is deduced from EAN rule.
EAN rules states – those complex’s in which effective atomic number is numerically equal to atomic number of noble gas element found in same period in which metal is situated, are most stable.
EAN (effective atomic number) =

A. [Co(CO)₄]^–
EAN = 27 – (–1) + 2 × 4
= 36 (inert gas atomic number)
Since, it already has EAN = 36 so it won’t give up or take in any electron.
B. EAN = 25 – (0) + 2 × 6 = 37
It has EAN = 37 so to attain inert gas electronic configuration it will donate one electron

It will act as reducing agent.
C. [Mn(CO)₅]
EAN = 25 – (0) + 2 × 5 = 35
To gain inert gas electronic configuration, it will accept one electron.

It will act as oxidizing agent.
D. [Cr(CO)₆]
EAN = 24 – (0) + 2 × 6 = 36
It already has inert gas configuration so it won’t exchange electrons.
Question 2
4 / -1

If P,Q,R and S are elements of 3^rd period of p–block in modern periodic table and among these one element is metal and rest are non-metal and their order of electronegativity is also given as
P < Q < R < S .Then in which of the following release of H⁺ is relatively easier.

A
P – O – H

B
S – O – H

C
Q – O – H

D
R – O – H

Solution

In any X – O – H, we take into account electronegative difference between X and O, and O and H. If electronegative difference between X and O is greater than O and H, X – O bond will be more polar and will break easily giving ^–OH. If electronegative difference between O and H is larger than X and O, O – H bond will be more polar and easier to break.
Electronegativity of S is largest, so electronegative difference between S and O will be least in S – O – H. It will be easier to break O – H bond, giving
Question 3
4 / -1

In which of the following ‘meta form’ of ‘-ic acid’ is not possible.

A
B(OH)₃

B
H₂SO₄

C
H₃PO₄

D
None of these

Solution

For meta form of an acid to exist it must be capable of giving one H₂O molecule and also contain at least one ‘H’ after donation of water.
Question 4
4 / -1

Bakelite is a

A
chain – growth polymer

B
step growth polymer

C
both a and b

D
elastomer

Solution
Question 5
4 / -1

If mechanism of reaction is

Where k is rate constant then, what is

A

B
K[A]

C

D
Kⁿ[A]

Solution
Question 6
4 / -1

which of the following species undergo non–redox thermal decomposition reaction on heating

A
FeSO₄

B
SnSO₄

C
H₂C₂O₄

D
Na₂HPO₄

Solution

Non–redox decomposition implies no change in oxidation number

On taking a look at these options we observe only in option (D) there has been no change in oxidation number.
Question 7
4 / -1

A

B

C

D

Solution
Question 8
4 / -1

k_sp for AgCl is-
Given: T = 25 °C

are 17.7, 13.2 and 23.0 cal/mol. (Antilog 0.21 =1.6)

A
1.6 × 10^–10

B
1.6 × 10^–9

C
1.6 × 10^–8

D
1.6 × 10^–7

Solution
Question 9
4 / -1

If NaCl is doped with 10^–2 mol% of SrCl₂, which of the following option shows concentration of cation vacancies?

A
10^–4 mol^–1

B
6.023 × 10¹⁹ mol^–1

C
6.023 × 10²¹ mol^–1

D
10⁶ mol^–1

Solution

Doping of NaCl with 10^–2 mol% SrCl₂ means that 100 mol of NaCl are doped with 10^–2 mol of SrCl₂.
100 mol of NaCl doped with = 10^–2 mol SrCl₂

Each ion introduces one cation vacancy,
therefore, cation vacancy: 10^–4 mol/mol of NaCl
= 10^–4 × 6.023 × 10²³ mol^–1
= 6.02 × 10¹⁹ mol^–1 of NaCl
Question 10
4 / -1

Product A is;

A

B

C

D

Solution