Assertion: During electrolysis of aqueous solution of NaCl2. Hydrogen is formed at anode and Oxygen is formed at cathode.
Reason: Ions gets attracted to oppositely charged electrons.

Dear Student,
First correct your question it is NaCl not NaCl2

The products of electrolysis of aqueous solution of NaCl can be explained as follow

• Electrolysis of aqueous  NaCl: 

Here, NaCl and Water H₂O both are present and both dissociate as

NaCl -----------> Na⁺ + Cl^-

H₂O--------------> H⁺ + OH^-

Reaction as cathode: Both Na⁺ and H⁺ will compete for cathode but reaction with higher E⁰ is preferred

 Na⁺ + e^- ----------> Na  E⁰ = -2.71 V  (1)

H⁺+ e^- -----------> 1/2 H₂   E⁰ = 0.00 V  (2)

Since, (2) has higher value of E⁰ So, H₂ is product deposited at cathode. 

Note-The eqn 2 can also be written as  H₂O + e^- --------------> 1/2 H₂ + OH^-

Reaction as anode: Both Cl^- and OH^- will compete for anode but reaction with lower E⁰ is preferred

Cl^-------------------> 1/2Cl₂ + e^-    E⁰ = +1.36 V    (3)

2H₂O --------------> O₂+ 4H⁺ + 4 e^-  E⁰ = +1.23 V  (4)

The eqn 4 is the reaction of OH^- at anode. The (4) has lower  E⁰, it should be preferred. But due to overpotential Cl₂ is preferred product at anode.

Hence,  the product of electrolysis are H₂ at cathode and Cl₂ at anode. 

Reason is true because it is ionic bond and Ions gets attracted to oppositely charged electrons.
Hence option D is correct answer

​​​​​​Hope it is clear
Regards