Due to the presence of a lone pair of electrons on N, NH3 acts as a complexing agent (ligand). As a result, it combines with transition metal cations to form complexes. For example:
\(AgCl\quad +\quad 2{ NH }_{ 3 }\longrightarrow \left[ Ag{ \left( { NH }_{ 3 } \right) }_{ 2 } \right] Cl\)
\(\\ Silver\quad chloride\quad \quad Diamminesilver\left( I \right) chloride\)
\(Cu{ SO }_{ 4 }\quad +\quad 4{ NH }_{ 3 }\longrightarrow \left[ Cu{ \left( { NH }_{ 3 } \right) }_{ 4 } \right] { SO }_{ 4 }\)
\(\\ Copper\quad sulphate\quad Tetraamminecopper\left( II \right) sulphate\)
\(Cr{ Cl }_{ 3 }\quad +\quad { 6NH }_{ 3 }\longrightarrow \left[ C{ r\left( { NH }_{ 3 } \right) }_{ 6 } \right] Cl_{ 3 }\) 
\(\\ Chromium\quad chloride\quad Hexaamminechromium\left( III \right) chloride\) ​​​​​

Electronic configuration of P is \(1{ s }^{ 2 }2{ s }^{ 2 }2{ s }^{ 6 }3{ s }^{ 2 }3{ p }_{ x }^{ 1 }3{ p }_{ y }^{ 1 }3{ p }_{ z }^{ 1 }3{ d }^{ 0 }.\) Thus, P has empty 3d orbitals to which the 3s

electron can be excited to have five half-filled orbitals needed for formation of \(P{ Cl }_{ 5 }.\) Thus, \(P{ Cl }_{ 5 }\) is known.
In contrast, electronic configuration of N is \(1{ s }^{ 2 }2{ s }^{ 2 }2{ p }_{ x }^{ 1 }2{ p }_{ y }^{ 1 }2{ p }_{ z }^{ 1 }.\) Since the valence shell of nitrogen has n = 2, therefore, it cannot have d-orbitals. However, if one of the 2s electron is excited to 3s orbital, five half-filled orbitals needed to form \({ NCl }_{ 5 }\) can still be obtained. But such an excitation is thermodynamically not favorable since the energy needed for excitation is more than the energy expected to be generated during the formation of two additional P - Cl bonds. Therefore, nitrogen does not form \({ NCl }_{ 5 }\). In other words, \({ NCl }_{ 5 }\) is unknown.

Dilute and conc.  Nitric acid gives different products of oxidation when reacted with copper metal.
3Cu(s) + 8HNO₃(dil.) \(\to\) 3Cu(No₃)₂(aq) + 2NO(g) + 4H₂O(l)
Cu(s) + 4HNO₃(conc.) \(\to\) Cu(NO₃)₂(aq) + 2NO₂ + 2H₂O(l)

(i) It is due to increases in size of group 15 elements, the bond angle decreases as bond length increases and bond pair-bond pair repulsion decreases.
(ii) It is because it is inert.

(i) XeF₆ + 3HO \(\to\) Xe + 2HF + \(1\over2\) O₂
(ii) \(​​\ \ \ {2NaOH} \\{ (cold\ and \ dill)}\) + Cl₂ \(\to\) NaCl + NaCIO + H₂O

(i)  it is because all the bonds are not in same plane.  Axial bonds are longer than equatorial bonds due to more repulsion.
(ii)  It is because of strong S-S bond than O-O due to greater repulsion between valence electrons of smaller atoms of oxygen than sulphur atoms.

(a) White phosphorus exists as discrete (separate) tetrahedral molecules. Thus, it has tetrahedral structure with six P-P bonds. Red phosphorus has polymeric  structure in which   molecules (tetrahedral) are linked through P-P bonds to form chain. Black phosphorus has two forms -black and -black phosphorus. It can be sublined in air and has opaque monoclinic or rhomohedral crystals; It does not oxidise in air. -black phosphorus does not burn in air up to 673 K. Black phosphorus is thermodynamically most stable.
Reactivity: White phosphorus is more reactive than red phosphorus because it is monomeric and has angular strain due to bond angles at 60°C. Black phosphorus is also less reactive.

(i) SbH₃ < AsH₃ < PH₃ < NH₃ is increasing order of basic strength.
(ii) HF < HCl < HBr < HI is increasing order of acidic strength.
(iii) HClO < HClO₂ < HClO₄ is increasing order of oxidising power.

(a) (i) It is because  \({ NF }_{ 3 }\) is more stable than \({ NCI }_{ 3 }\)because \({ F }_{ 2 }\)is stronger oxidising agent than \({ CI }_{ 2 }\) .
(ii) It is due to low bond dissociation energy, high hydration energy, high electron affinity.
(b) \((i)\ C+{ 2H }_{ 2 }{ SO }_{ 4 }(conc.)\longrightarrow { CO }_{ 2 }+{ 2SO }_{ 2 }+2H_{ 2 }O\)
\(\\ (ii)\ { P }_{ 4 }+3NaOH+3H_{ 2 }O\longrightarrow { 3NaH }_{ 2 }{ PO }_{ 2 }+PH_{ 3 }\)
\(\\ (iii)\ { CI }_{ 2 }+3{ F }_{ 2 }\longrightarrow 2CIF_{ 3 }\)

(a)\((i)\quad { 2I }^{ - }+{ O }_{ 3 }+{ H }_{ 2 }{ O }\longrightarrow { I }_{ 2 }+{ O }_{ 2 }+2{ OH }^{ - }\)
\(\\ (ii)\quad { P }_{ 4 }(s)+3NaOH(aq)+{ 3H }_{ 2 }{ O }(l)\longrightarrow { PH }_{ 3 }{ (g) }+{ 3NaH }_{ 2 }{ PO }_{ 2 }\)
(b)(i) It is because bond dissociation energy of H-Te bond is less than H-S bond due to longer bond length.
(ii) It is due to higher standard reduction potential, low bond dissociation energy, high electron affinity and higher enthalpy of hydration.
(iii) It is due to stable electronic configuration i.e., their octet is complete except in He which has duplet i.e., 1st shell is complete having 2 electrons.

N does not have d-orbitals in the valence shell