Chemistry Test - 4

\(\left[\mathrm{Cu}(\mathrm{en})_{2}\right]^{+2}\)can not exhibit geometrical isomerism.

Complex in which there is symmetry, do not possess geometry isomerism. Complex \(\left[\mathrm{Cu}(\mathrm{en})_{2}\right]^{+2}\) has bidentate ligend and has symmetry. Thus do not show geometrical isomerism.

1) Even though the aqueous hydrogen halides are themselves strong acids, the presence of additional sulphuric acid speeds up the formation of alkyl halides.

So, option (1) is correct.

(2) The rearrangement of alkyl groups occurs, except with most primary alcohols.

So, option (2) is wrong.

(3) The order of reactivity of alcohols towards HX is 30 >20 >10< CH₃. The reactivity decreases through a minimum at 10 and rises again at CH₃.

So, option (3) is also wrong.

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

\(N H_{3} \rightarrow N H_{2}^{*}+H^{+}\)

Water being more polar than ammonia dissociates more and gives plenty amount of \(H^{+}\) and hence \(H_{2}\) gas is released as result. In ammonia sufficient amount of \(H^{+}\) is not obtained due to less polar nature (so less dissociation) hence \(H_{2}\) gas is not evolved and then ammoniation of electron happens

b) As we dilute further dissociation of ammonia increases (because degree of dissociation is inversely proportional to concentration) and we get more \(H^{+}\) and \(N H_{2}\) ion. \(H^{+}\) combines with ammoniated electron and \(H_{2}\) gas is evolved and paramagnetism is gone.

c) \(N H_{3} \rightarrow N H_{2}+H^{+}\)

\(M \rightarrow M^{+}+e^{-}\)

\(M^{+}+x N H_{2}^{-} \rightarrow\left[M\left(N H_{2}\right) x\right]\) type of complex

On adding d block metal ion we see that metal ion combines with \(N H_{2}\) and forms metal complex. Complex formation leads to ammonia dissociation even more and as a result of ammonia dissociation more \(H^{+}\) is released and which combines with ammoniated electron and \(H_{2}\) gas is evolved. Now colour and magnetic nature may or may not change. But whatever it will be it will be due to this complex.

d) Metallic bonding increases at very high concentration \(M_{2}\) is formed.

Hence option (d) is correct.

Condition for ‘ligand to metal charge transfer’ are

i) Metal should be in high oxidation state so that it has high ionization energy. And it should also be of smaller size with vacant orbitals having low energies.

ii) Ligand should have lone pair of electrons having high energy

In KMnO₄, Mn is in +7 oxidation state and have all the 3d orbitals vacant. Mn⁺⁷ ion is surrounded by four oxide ions. All oxide ions have filled 2p orbitals. There is transfer of an electron from filled 2p orbital of oxide ion to vacant d orbitals of Mn⁺⁷ ion. Due to this transfer of electron from oxides of KMnO₄ to metal happens and as a result colouris intensely purple.

Trick: Almost all the metal ion of d block having d0 configuration tend to show colour due to ligand to metal charge transfer. Another example of this is K₂Cr₂O₇

Hence option (A) is correct.

1. \(M g^{2+}+2 e^{-} \rightarrow M g\)

2F deposits 1 mole of \(\mathrm{Mg}\) 1F deposits 0.5 mole of \(M g\)

Thus, option (1) is correct.

2. \(A l^{3+}+3 e^{-} \rightarrow A l\)

\(3 F\) deposits 27 mole of \(M g\) \(1 \mathrm{F}\) deposits \(\left(\frac{27}{3}\right)=9\) mole of \(A L\)

Thus, option (2) is also correct.

\(4.4 O H \rightarrow 2 H_{2} O+O_{2}+4 e^{-}\)

3 F liberates \(22.4 L\) of \(O_{2}\) at STP \(1 \mathrm{Fliberates} \frac{22.4}{4}=5.6 \mathrm{LofO}_{2}\)

Thus, option (3) is also correct.

(a )(NH₄)₂Cr₂0₇→ N₂ + Cr₂O₃ +4H₂O

b) 8NH₃ + 3Br₂→N₂ + 6NH₄Br

c) 2NH₃ + 3CuO →N₂ + 3Cu + 3H₂O

d) 3Ca + 6NH₃→3[Ca(NH₂)₆]

Hence option (d) is correct.

Hydrogen ₁H : 1S¹

Hydride: ₁H- :1S²

Helium ₂He : 1S²

Helium has two proton in it and it can stabilize 2 electrons. But, in hydrogen as one more electron enters and it becomes hydride ion, now one proton has to account for two electrons, the stability is affected. In hydride one proton cannot stabilize two electrons so structural deformity arises which leads to its reactivity.

Hence option (b) is correct.

1. Guanidine is a stronger base than pyridine. The increased basicity can be explained by drawing the resonance structures of the protonated guanidine.

2. Dimethylamine is more basic than trimethylamine in aqueous medium. This is due to the fact that steric

hindrance created by alkyl groups affect the solvation so badly that the tertiary amines are usually the least basic among the three classes. Due to the presence of 3 alkyl groups, there is a lot of steric hindrance and therefore, it is difficult for nitrogen to form H-bonds with the hydrogen atom of water for proper solvation.

3. 2,4,6 -Trinitro-N, N-dimethyl aniline is a stronger base than 2,4,6 -Trinitroaniline because the steric repulsion of -

\(N\left(C H_{3}\right)^{2}\) goes out of the plane of the ring.

Hence option D is the correct answer.

Reaction mentioned is given as below

2XeF₂ +2H₂0 →2Xe + 4HF + O₂

HF forms H^-bond with F^- and exists as HF^2-

Hence option (d) is correct.

PVC (polyvinyl chloride) and polythene are thermoplastics.

Thermoplastics: Thermoplastics can be identified as those plastics polymers which are softened by heating and hardened by cooling. This can simply be transliterated as these polymers have the capacity of changing shape and form if sufficient amount of heat is supplied to the system

Thermosetting plastics: Thermosetting plastics have the exact opposite characteristics when compared to thermoplastics. This means that these plastic polymers do not change shape or form when heat is provided to the system. Instead, they tend to burn out in the process.

Natural Substances: Natural substances are those which are obtained directly from nature and are used with little or no processing.

Natural Fibres: Natural Fibres are a part of natural substances. These fibres require some processing before being put into direct use.

Now let us discuss the given polymer plastics −PVC

−PVC and polythene. Both of these plastic polymers are obtained by the chemical processing of different monomers. Hence, they are not naturally available substances. As for the discussion whether they are thermosetting or thermoplastic, we can deduce this firm fact that they change shape and form upon heating.

So, PVC (polyvinyl chloride) and polythene are thermoplastics.

α-amino acids have the structure

If \(R \neq H,\) the central carbon atom is asymmetric. So, the compound will be optically active. Thus, (1) is true.

For glycine, \(R=H . S O,\) it is optically inactive. So, (2) is true.

All natural \(\alpha\) -amino acids have L-configuration as shown below.

since these contain an acidic (-COOH) group and a basic (-NHa) group, they behave neither as completely acidic nor as completely basic. Thus, they are amphoteric in nature.

So, (3) is correct.

a) \(B(O H)_{3}+2 H_{2} O \rightarrow B(O H)_{4}+H_{3} O^{+}\)

This reaction is characterized as Lewis Acidity of boric acid

b) Due to hydrogen bond present in crystalline structure it becomes difficult for it to donate \(H+\) and hence acidic character decreases.

c) \(B(O H)_{3}+2 H_{2} O \rightarrow B(O H)_{4}+H_{3} O^{+}\)

Cis-diols form complex with borate ion and shift the equation forward hence ionizing the boric acid to higher extent. And acidic character increases.

Hence option (c) is correct.

Balanced Reaction is :

2Ca₃(PO₄)₂ + 6 SiO₂ +10 C → 6CaSiO₃ + 10CO + P₄

Hence option (d) is correct.

As soon as Cu(CN)₂ or CuI₂ formed, due to big size of CN^- and I^-, they become very unstable. Because big size of anion leads to steric hindrance and they decompose into CuCN and CuI.

Hence option (c) is correct.

a) In graphite there is delocalization of electron along the layer and it is good conductor. Across the layer graphite and good enough gap to diminish the electron movement and it acts as semiconductor. On increasing the temperature conduction of conductors decreases and semiconductors increases.

Graphite

b) Graphite is thermodyanamically more stable because there are conjugate double bond. But diamond is kinetically more stable.

c) Buckminster fullerene can trap lanthanum and other metals of appropriate size in its cavity.

d) Diamond doesn't has any double bond.

Hence option (c) is correct.