Chemistry Test - 38

Primary valency is ionizable according to Werner's theory of coordination compounds.

According to Werner's theory, a coordination compound has two different types of valency, primary and secondary. The primary valency or the ionizable valency is satisfied by the negatively charged ions in the solution.

Primary valencyis ionizable and is satisfied by the negative charges whereasSecondary valency is non-ionizable and is satisfied by the positive charged or neutral species in the solution.The primary valency corresponds to the oxidation state of the metal ion. The secondary valency corresponds to the coordination number of the metal complex. The molecules or ions that satisfy the secondary valency are called ligands and they can be either negatively charged or neutral.

For example, in \(\left[ Cu \left( NH _{3}\right)_{4}\right] SO _{4}\) primary valency is 2 and secondary valency is 4. Secondary valence refers to coordination number. Since copper is coordinated to 4 ammonia ligands, secondary valence is 4. Primary valence is satisfied by anions. Since sulphate ion has \(-2\) charge, primary valence is 2.

Correctly match functional group given in List-I with the Nomenclature of that functional group given in list-II:

(a) \(\mathrm{B}_2 \mathrm{O}_3 \rightarrow\) oxide of boron (non-metal) : acidic

\(\mathrm{CaO} \rightarrow\) oxide of calcium (metal) : basic

(b) \(\mathrm{B}_2 \mathrm{O}_3 \rightarrow\) oxide of boron (non-metal) : acidic

\(\mathrm{SiO}_2 \rightarrow\) oxide of silicon (non-metal) : acidic

(c) \(\mathrm{N}_2 \mathrm{O} \rightarrow\) oxide of nitrogen (non-metal) : neutral

\(\mathrm{BaO} \rightarrow\) oxide of barium (metal) : basic

(d) \(\mathrm{CaO} \rightarrow\) oxide of calcium (metal): basic

\(\mathrm{SiO}_2 \rightarrow\) oxide of silicon (non-metal): acidic

So, option (b) is the correct answer. \(\mathrm{B}_2 \mathrm{O}_3\) and \(\mathrm{SiO}_2\) are being acidic oxide they react with base to give salt and water.

\(\mathrm{B}_2 \mathrm{O}_3+2 \mathrm{NaOH} \rightarrow \underset{\text{Sodium metaborate}}{2 \mathrm{NaBO}_2}+\mathrm{H}_2 \mathrm{O} \)

\(\mathrm{SiO}_2+2 \mathrm{NaOH} \rightarrow \underset{\text{Sodium silicate}}{\mathrm{Na}_2 \mathrm{SiO}_3}+\mathrm{H}_2 \mathrm{O} \)

Carbon belongs to the second period and Group 14. Silicon belongs to the third period and Group 14. If atomic number of carbon is 6, the atomic number of silicon is 14.

Silicon is a chemical element with the symbol Si and atomic number 14. It is a hard, brittle crystalline solid with a blue-grey metallic lustre, and is a tetravalent metalloid and semiconductor. It is a member of group 14 in the periodic table: carbon is above it; and germanium, tin, lead, and flerovium are below it.

Presence of one vinyl (or ethenyl) group gives formaldehyde as one of the product in ozonolysis.

CH₂=CH- on ozonolysis will give HCHO.

• Ozonolysis is an organic reaction where the unsaturated bonds of alkenes, alkynes, or azo compounds are cleaved with ozone. 
• Alkenes and alkynes form organic compounds in which the multiple carbon–carbon bond has been replaced by a carbonyl group while azo compounds form nitrosamine.

\(Q = i \times t\)

\(Q =10 \times 10^{-3} \times t\)

\(2 H_{2} O+2 e^{-} \rightarrow H_{2}+2 O H^{-}\)

To liberate \(0.01\) mole of \(H_{2}\), \(0.02\) Faraday charge is required

\(Q=0.02 \times 96500 C\)

\(\therefore 0.02 \times 96500=10^{-2} \times t\)

\(\Rightarrow t=19.3 \times 10^{4}\) sec

\( \text { Applying ; }\left(n_{\mid}+n_{||}\right) \text {initial }=\left(n_{\mid}+n_{\|}\right)_{\text {final }} \)

\( \Rightarrow \text { Assuming the system attains a final temperature of } T \text { (such that } 300< T <60) \)

\( \Rightarrow\left(\begin{array}{c}\text { Heat lost by } \\ N _2 \text { of container } \\ \text { I }\end{array}\right)=\left(\begin{array}{c}\text { Heat gained by } \\ N _2 \text { of container } \\ \text { II }\end{array}\right) \\ \)

\( \Rightarrow n _1 C _{ m }(300- T )= n _{ n } C _{ m }( T -60) \)

\( \Rightarrow\left(\frac{2.8}{28}\right)(300- T )=\frac{0.2}{28}( T -60) \\ \)

\( \Rightarrow 14(300- T )= T -60 \)

\( \Rightarrow \frac{(14 \times 300+60)}{15}= T \)

\( \Rightarrow T =284 K \text { (final temperature) } \)

\( \Rightarrow \text { If the final pressure }= P \)

\( \Rightarrow\left( n _1+ n _{\text {II }}\right)_{\text {final }}=\left(\frac{3.0}{28}\right) \\ \)

\( \Rightarrow \frac{ P }{ RT }\left( V _{ I }+ V _{ II }\right)=\frac{3.0 gm }{28 gm / mol } \)

\( P =\left(\frac{3}{28} mol \right) \times 8.31 \frac{ J }{ mol - K } \times \frac{284 K }{3 \times 10^{-3} m ^3} \times 10^{-5} \frac{ bar }{ Pa } \)

\( \Rightarrow 0.84287 bar \)

\( \Rightarrow 84.28 \times 10^{-2} bar = 84\)

\(\mathrm{O}_2{ }^{+}(15)\)

\(=\mathrm{KK} \sigma 2 \mathrm{~s}^2 \sigma 2 \mathrm{~s}^2 \sigma 2 \mathrm{p}_{\mathrm{z}}{ }^2 \pi 2 \mathrm{p}_{\mathrm{x}}{ }^2=\pi 2 \mathrm{p}_{\mathrm{y}}{ }^2 \pi^3 2 \mathrm{p}_{\mathrm{x}}{ }^1=\pi 2 \mathrm{p}_{\mathrm{y}}{ }^0\)

\(\text { Bond order }=\frac{1}{2}(8-3)=\frac{5}{2}=2.5 \)

\(\mathrm{O}_2(16)=\mathrm{K} \mathrm{K} \sigma 2 \mathrm{~s}^2 \sigma^{\mathrm{t}} 2 \mathrm{~s}^2 \sigma 2 \mathrm{p}_{\mathrm{z}}{ }^2 \pi 2 \mathrm{p}_{\mathrm{x}}{ }^2=\pi 2 \mathrm{p}_{\mathrm{y}}{ }^2 \pi^* 2 \mathrm{p}_{\mathrm{x}}{ }^1=\pi^* 2 \mathrm{p}_{\mathrm{y}}{ }^1 \)

\(\text { Bond order }=\frac{1}{2}(8-4)=2 \)

\( \mathrm{O}_2{ }^{-}(17)=\mathrm{KK} \sigma 2 \mathrm{~s}^2 \sigma^2 2 \mathrm{~s}^2 \sigma 2 \mathrm{p}_{\mathrm{z}}{ }^2 \pi 2 \mathrm{p}_{\mathrm{x}}{ }^2=\pi 2 \mathrm{p}_{\mathrm{y}}{ }^2 \pi^* 2 \mathrm{p}_{\mathrm{x}}{ }^2=\pi^* 2 \mathrm{p}_{\mathrm{y}}{ }^1 \)

\(\text { Bond order }=\frac{1}{2}(8-5)=1.5\)

\(\mathrm{O}_2{ }^2(18)=\mathrm{KK} \sigma 2 \mathrm{~s}^2 \sigma^* 2 \mathrm{~s}^2 \sigma 2 \mathrm{p}_{\mathrm{z}}{ }^2 \pi 2 \mathrm{p}_{\mathrm{x}}{ }^2=\pi 2 \mathrm{p}_{\mathrm{y}}{ }^2 \pi^* 2 \mathrm{p}_{\mathrm{x}}{ }^2=\pi^* 2 \mathrm{p}_{\mathrm{y}}{ }^2 \)

\(\text { Bond order }=\frac{1}{2}(8-6)=1 \)

Note: As we know that as the bond order decreases, stability also decreases and hence the bond strength also decreases. Therefore the correct order of their increasing bond strength is

\(\mathrm{O}_2{ }^2<\mathrm{O}_2{ }^{-}<\mathrm{O}_2<\mathrm{O}_2{ }^{+}\)

Given that:

Atomic Number, \(({Z} )=27\)

The configuration will be:

\( [{Ar}] 3 {~d}^{7} 4 {~s}^{2}\)

\(\therefore {M}^{2+}=[{Ar}] 3 {~d}^{7}\)

\(3 {~d}^{7}:\)  

i.e., 3 unpaired electrons,

\(\therefore {n}=3\)

We know that, 

Magnetic moment is given by:

\( \sqrt{n(n+2)}=\mu\)

where, \(\mathrm{n}=\) total number of unpaired electron

\( \sqrt{3(3+2)}=\mu\)

\( \sqrt{15}=\mu\)

\(\mu =3.8\approx 4 {~BM}\)