In the first equation electron is added to a gaseous anion and energy is released in this process and in the second case also energy is released .

Both 1 and 2 are correct because , the electronic configuration of Au is given by  = Au = 4d¹⁰ 4f¹⁴ 6s¹.  The electronic configuration of Ag is given by = Ag = 4d¹⁰5s¹ . The electronic configuration of Cu is given by = Cu = 3d¹⁰ 4s¹   . The effective nuclear charge of Au increases  because of  poor shieiding effect  of 4f¹⁴  electrons (  lanthanoid contraction ) , due to which its ionization energy is higher in its group .  electronic configuration of Cu is given by  - 3d¹⁰4s¹ .  electronic configuration of K is given by - 4s¹3d⁰ . In Cu poor shielding effect of  3d¹⁰ electrons causes increase in effective nuclear charge and because of this IP value of Cu is greater than that of K . Hence, both the options are correct.

 If abigger anion is attached to the alkali and alkaline earth metal groups , solubility decrease down the group . because solubilty depends on the the lattice energy  and hydration energy and in this case lattice energy > hydration energy . U _{lattice energy}  = ( Z⁺ Z^-) / (r ⁺ + r^-) and  U _{hydration energy}  = (Z⁺/r⁺) +( Z^-/r^- ) . Lattice energy increases down the group at faster rate than that of hydration energy . Hence, this option is correct.

Polarisibilty is given by = (anionic charge) ( anionic charge) . charge of N , O , F  is same i.e -1 but the radius is in the order of  N > O > F. Hence, this option is correct . 

The electronic configuration of Cu is given by  - 3d¹⁰ 4s¹ , therefore Cu⁺ has 3d¹⁰ 4s⁰ . The electronic configuration of K is given by  - 4s¹ 3d⁰ . and therefore K⁺ has  4s⁰ 3d⁰ . In Cu pseudo noble gas configuration is there because of this ionic potential of of Cu increases. It has more covalent character . Hence, less melting point . 

This option is correct. In Pb⁺² inert pair effect is there i.e. because of lanthanoid contraction , poor shieding effect of  4f¹⁴  electrons . The effective nuclear charge increases and after removal of two  electrons it becomes really hard to remove the other two electrons . Hence, Pb⁺² is most stable among its group because lanthanoids contraction observe in this atom only .

The conductivity depends on the value of ionic potential . Higher the value of ionic potential lower is the conductivity. Ionic potential = charge of cation / radius of cation . µ_Be>µ_Mg>µ_Ca>µ_Ba . Hence, this option is correct .

this option is correct because , through  molecular orbital theory , we get the bond order of CO to be 3 . and the last electron lies  in the bonding molecual orbital but actually energy of sigma  antibonding 2s²  orbital is greater than that of sigma 2p orbital . hence the removal of electron takes place from  sigma  antibonding 2s²  orbital .  which increase the bond order to 3.5  which is calculated by the formula  1/2 (N_b- N_a) , where N_b and N_a are the number of electrons in bonding and anti - bonding orbitals respectively . 

F₂ > Cl₂ > Br₂ > I₂  . hence the order of acidity is  SiF₄ > SiCl₄ > SiBr₄ > SiI₄ .

In protic solvent higher the value of ionic potential , highly hydrate  the cation is and therefore lower will be the mobilty . The ionic potential order is given by - Li+ > Na⁺ > K⁺ > Rb⁺ , therefore order of mobility is  - Rb⁺ > K⁺ > Na⁺ > Li⁺ .

This option is correct because , in ClOH chlorine has three lone pair and the electron on O^- is unable to delocalise itself . Hence, it will remain on oxygen atom and  has the highest basicity among the given options .  

This option is correct because , during the hydrolysis ,  water being neucleophile always prefer to attack the central atom first . it will attack the central atom  only when it posses vacant orbitals . If the central atom doesn't posses vacant orbital then attack  of water molecules taken place on peripheral atom provided they also posses vacant orbitals . in this case niether central nor peripheral atom posses vacant orbital hence hydroysis doesn't takes place in this molecule . 

The electronic configuration of N is 1s² 2s² 2p³ . 's' and 'p' orbital of nitrogen atom requires energy in order to for the hybrid orbital . and after that it will combine with three hydrogen atoms in which energy is released which is greater than  the energy requires for the formation of hybrid orbital . If this condition is satisfies than only bond is formed . but in case of  phosphorus and higher elements these compensation of energy is not that much which compensate the energy of hybridization . hence each P orbital overlap individually & hence there bond angle  is around 90 degree and whatever fluctuation about 90 degree is because of replusion . As we move down the  group electron will be then donated form pure 's' orbital . Hence,  the correct order is BiH₃ <  SbH₃ < AsH₃ < PH₃ < NH₃ .

This option is correct. Bond order of N₂ = 3. Bond order of N₂⁺ = 2.5. Bond order of N₂^- = 2.5. Since bond order of N₂⁺ and N₂^- is the same, but the effective nuclear charge of N₂⁺ is more than that of N₂^-, so the correct order is N₂ > N₂⁺ > N₂^-.

This option is correct because in the case of CH₃F, magnitude of charge is the highest, but in the case of CH₃Cl, the charge is almost the same as that of CH₃F, but the separation between the charges overpowers. Hence, the magnitude of dipole moment of CH₃Cl is greater than that of CH₃F. Hence, the correct order is CH₃I < CH₃Br < CH₃F < CH₃Cl. 

You might have remembered  this logic  , if the total number of electrons comes out to be 50  the hybridisation is capped octahedral i.e. sp³d² or distorted octahedral .  

this option is correct. In Cl₂O ,  the two lone pair of electrons on Oxygen atom get associated in making of  Pπ - d π back bonding with the empty 'd' orbital of Cl  and this replusion is higher as compared to other options . In the case of H₂O ,  oxygen is more electronegative than hydrogen. Hence,  the  shared pair of electrons will lie more closer to oxygen than hydrogen and  the bond pair - bond pair replusion is there which is more than the replusion between the electrons density of two florine atoms in case of Fl₂O . 

 this option is correct because , in BF₃ boron has empty 'p' orbital in which non bonded electrons of Fluorine accommodate and form pπ - pπ back bond . due to which bond length decreases . hence  this statement  is incorrect . and  this  option is correct  . 

this option is correct because , in case of homoneuclear molecule it is easy to calculate the magnetic character through molecular orbital theory . but in case of hetronuclear molecule one  easy method is there to find out the magnetic character . magnetic character in case of hetronuclear molecules . 1. count  the total number of valence electrons if it comes out to be even then it is dimagnetic otherwise paramagnetic . in this case Cl has 7 valence electrons and oxygen has 6 valence electrons . so total number of valence electrons is 7 + 6 x 2 = 19 , which is odd hence this molecule is paramagnetic . and this option is correct  .   

pπ- pπ back bonding is there  in Cl due to which partial double bond character is there and hence, it is higher in order , but in case of bromine partial double bond, character is there but because of its large size double bond character is less than that in the case of Cl₂ . This double bond character is not observe in case of iodine because of its large size . Florine has smaller in size than  iodine therefore it is above the iodine in order . Hence, the correct order  is Cl₂ >  Br₂ > F₂ > I₂ .