Chemistry Test - 20

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 KMnO_4, 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.

a) H₂O →OH- + H⁺

NH₃→NH₂^- +H⁺

Water being more polar than ammonia dissociates more and gives plenty amount of H⁺ and hence H₂ gas is released as result. In ammonia sufficient amount of H⁺ is not obtained due to less polar nature (so less dissociation) hence H₂ 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 NH₂^- ion. H⁺ combines with ammoniated electron and H₂ gas is evolved and paramagnetism is gone.

c) NH₃→NH₂^- +H⁺

M →M⁺ +e^-

M⁺ +xNH₂^-→[M (NH₂)x] type of complex

On adding d block metal ion we see that metal ion combines with NH₂^- 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 H₂ 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 M₂ is formed.

Hence option (D) is correct.

Balanced Reaction is :
2Ca₃(PO₄)₂+ 6 SiO₂ +10 C → 6CaSiO₃ + 10CO + P₄
Hence option (D) is correct.

Reaction mentioned is given as below
2XeF₂ +2H₂0 →2Xe + 4HF + 0₂
HF forms H-bond with F^- and exists as HF₂^-
Hence option (D) is correct.

As soon as Cu(CN)2or CuI2formed, 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.

IUPAC name of the compound will be

2,7-dimethyl-3,5-octadiyne-2,7-diol

Hence option (A) is correct.

Position a and c will be more susceptible to nucleophilic attack
Position a has bromine attached to it making it electron deficient or more electropositive and hence inviting for nucleophile
Position c has oxygen atom attached to it. High electronegativity of oxygen makes carbon highly electropositive and susceptible to nucleophilic attack.
Hence option (C) is correct.

Mechanism of the reaction

1. BuNH₂ acts as nucleophile and attacks on conjugate double bond and releases OEt^-

2. OEt^- acts as base now and removes the proton from nitrogen

3. Then EtOH formed will provide the proton to double bond

Hence option (C) is correct.

In water the product is almost all benzyl naphthol. However, in DMSO (dimethyl sulfoxide) the
major product is the ether. In water the oxyanion is heavily solvated through hydrogen bonds to
water molecules and the electrophile cannot push them aside to get close to O– (this is an entropy
effect). DMSO cannot form hydrogen bonds as it has no OH bonds and does not solvate the oxyanion, which is free to attack the electrophile.

Anticlockwise direction implies that angel will be taken negative.
10cm path length means 1 dm.
28mg in 1cm3 is given we need to change it to g/cm3= .028g/cm3
Specific Rotation = -4.35/.028 = -155.35