How to deal with chemguys practice set- 1?

Chemguys practice Set -1: (share👇👇🏻👇👇👇📝📝📚📚)

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Content :
1. Practice set -1
→ 8 MCQ
→ 2 NAT
2. Answer key of practice set.
3. Explanation of all questions
4. Self-test ( check your concept that you have learned while solving practice set)
5. Answer key of self-test.

Practice Set:

Q.1:

The reagent suitable for effecting the following transformation is

a) CH₂N₂
b) Me₂CuLi
c) Meli
d) Ph₃P=CH₂

Q.2:

Which of the following thermodynamic relation is correct?
a ) dG=VdP-SdT
b ) dE= PdV+TdS
c)dH= -VdP+TdS
d) dG=VdP+SdT

Q.3:

The hybridization in SF6 molecule is :
a) sp³d²
b) sp²d³
c) spd⁴
d) sp³s¹p¹
Q.4:
What is the correct statement of the following two species?

H₃O⁺ and NO₃^-

a) similar in the hybridization of a central atom with different structures
b)H₃O⁺ has sp³ hybridization of the central atom and tetrahedral shape
c) NO₃^- has sp hybridization of the central atom.
d)none.

Q.5:

Using Wade's rule, the structure of the B₁₀C₂H₁₂ will be
a) Nido
b) close
c) arachnoid
d) none of these.

Q.6:

Arrange in order of decreasing acidic strength

a) X>Y>Z
b) X>Z>Y
c) Z>Y>X
d) Y>Z>X

Q.7:

For which order reaction does the unit of the rate equals the team of the rate constant?
a) 0
b) 1
c) 2
d) 4

Q.8:

The cubic unit cell of a metal ( molar mass = 63.55 g.mol^-1) has an edge length of 362 pm. Its density is 8.92 g.cm -³, the no. of atoms present in unit cell is ______________.

Q.9:

In the reaction of sodium thiosulphate with I₂ in an aqueous medium, the equivalent weight of sodium thiosulphate is equal to _________.

Q.10:

Answer key:
1 ) d , 2) a , 3 ) a , 4 ) d , 5 ) b , 6 ) b , 7 ) a , 8 ) 4 , 9 ) Equivalent weight of sodium thiosulphate , 10 ) D
Answer with Explanation:

Q.1:
This is an example of Wittig's reaction. The transformation and mechanism are as follows:

Q.2:

We know that,
Gibbs free energy,
G = H - TS
So, dG = dH -TdS - SdT ..............(i)
Now, enthalpy , H = U + PV
So, dH = dU + PdV + VdP
             = dQ + VdP [ From first law thermo
                                     - dynamics ]
            = TdS + VdP [ 2nd law of thermo
                                     - dynamics ]
Substituting the ' dH' in equation (i) , we get

dG = TdS + VdP - TdS - SdT = VdP - SdT

So the correct answer is a.

Q.3:

SF₆:
Hybridization=( 1/2)[ no of valence shell electrons + no of monoatoms - no of positive charge + no of negative charge ]
So, H_(SF6)
= ( 6+ 6 -0 + 0) / 2
= 6
This number means that sp³d².

Hybridization table:

Sp	2
Sp2	3
Sp3	4
Sp3d	5
Sp3d2	6

Q.4:

We know that,

Hybridization =( 1/2)[ no of valence shell electrons + no of monoatomic - no of positive charge + no of negative charge ]
No of lone pair = H - no of monoatomic - no of diatoms.
H₃O⁺:
Here O is the central atom.
H= (1/2)[ 6 - 3 +1] = 4

This means sp³.
Now, no of lone pair = 4 - 3 -0 =1
So, the structure will be pyramidal.
NO₃^-:
H = (1/2) [ 5 +1] = 3
3→sp²
Lone pair = 3 - 0 - 3 = 0
So, the structure will be Trigonal planar .

So, this does not match with given option i.e. answer will be option D .

Q.5:

B₁₀C₂H_{12 (}given) is a carboborane .
So, to convert it into Borane type, we replace C = BH.
So, what will we get now?
=> B₁₀C₂H₁₂ = B₁₀(BH)₂H₁₂ = B₁₀B₂H₂H₁₂ = B₁₂H₁₄
This leads to ths following type B_nH_n^2-
and that's why it is closo one.

Q.6:

Keep in mind the following points to find out the acidity order X, Y, and Z.
i) acidity ∞ stability of the conjugate base
ii) acidity ∞ easiness of proton loss.

Q.7:

The unit of the rate constant of n-th order is mol^{1-n lint}^-1 S^-1
Again, rate = dc/DT, so unit will be mol lit^-1 S^-1
So, it is clear from the above two units that 1- n = 1 & n - 1 = -1
That is n = 0 . So, the order of the reaction is Zero.

Q.8:

Q.9:

Q.10:

The checking factors for this question are stability of carbocation that formed in the middle of reaction that predicts (=O ) position and the stereochemistry of H atom migration.
In [1,2] rearrangements, the migrating H atom always is connected to the same face of the carbenium ion from which it begins the migration.

It follows that the [1,2]-shift in the carbenium ion B must proceed stereoselectively. This is because the H atom can begin its migration only on one side of the carbenium ion.

This rearrangement is stereoselective since there is only one H atom next to the sextet center and the H atom undergoes the[1,2]-migration on the same face of the five-membered ring.

Self-test

(Check your concept that you have learned through the above questions)

Q.1:

C₂B₉H₁₁ + 2e^- → A
A will be
a) close
b) Nido
c) arachnoid
d) Hypo
Q.2:
Wittig reaction goes through a
a) 3 membered cyclic intermediate
b) 4 membered cyclic intermediate
c) 6 membered cyclic intermediate
d) no cyclic intermediate formed.