Enthalpy is not a conclusive measurement of whether a reaction will be more spontaneous and favorable. Entropy is.

Taking a look at both reactions, in reaction A two molecules are combining and giving one molecule and in reaction B only one molecule is giving one molecule so decrease in entropy in reaction B is less. So, ∆S will be less negative in here hence more favorable.

ΔG = ΔH-TΔS

Hence option (d) is 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.

IUPAC name of the compound will be

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

Hence option (a) is correct.

Reaction mentioned is given as below

2XeF₂ +2H₂O →2Xe + 4HF + O₂

HF forms H-bond with F^- and exists as HF₂^-

Hence option (d) 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.

E₂ eliminations therefore take place from the anti-periplanar conformation.

Here, E₂ elimination gives mainly one of two possible stereoisomers.

2-Bromobutane has two conformations with H and Br anti-periplanar, but the one that is less hindered leads to more of the product, and the E-alkene predominates.

Another way to understand will be in any alkene formation  major product is always most stable alkene. We can see alkene given in option a is trans alkene and in option b is cis alkene. Trans alkene being more stable on account of less steric repulsion will be major product.

Hence option (a) is correct.

Here we can see that on increasing temperature those reaction having high value of E_a will be more susceptible to change in rate.

Hence, option (b) is correct.

a) B(OH)₃ +2 H₂O →B(OH)₄^- +H₃O⁺

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(OH)₃ +2 H₂O →B(OH)₄^- +H₃O⁺

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.