If four different elements A, B, C and D having outer electronic configuration as,
\(\text{A} = 3s^2 3p^4, \text{B} = 3s^2 3p^5, \text{C} = 4s^2 4p^4, \text{D} = 4s^2 4p^5\)
identify the element with lowest ionization enthalpy \((\Delta_i \text{H}_1)\)

Question

Given below are two statements :
Statement I: The boiling point of hydrides of Group 16 elements follow the order
H₂O > H₂Te > H₂Se > H₂S.
Statement II: On the basis of molecular mass, H₂O is expected to have lower boiling point than the other members of the group but due to the presence of extensive H-bonding in H₂O, it has higher boiling point.
In the light of the above statements, choose the correct answer from the options given below :

If four different elements A, B, C and D having outer electronic configuration as,
\(\text{A} = 3s^2 3p^4, \text{B} = 3s^2 3p^5, \text{C} = 4s^2 4p^4, \text{D} = 4s^2 4p^5\)
identify the element with lowest ionization enthalpy \((\Delta_i \text{H}_1)\)

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The Correct Option is C

Solution and Explanation

Top Questions on Group 16 Elements

Questions Asked in MHT CET exam

If four different elements A, B, C and D having outer electronic configuration as, \(\text{A} = 3s^2 3p^4, \text{B} = 3s^2 3p^5, \text{C} = 4s^2 4p^4, \text{D} = 4s^2 4p^5\) identify the element with lowest ionization enthalpy \((\Delta_i \text{H}_1)\)

Show Hint

The Correct Option is C

Solution and Explanation

Top Questions on Group 16 Elements

Questions Asked in MHT CET exam

If four different elements A, B, C and D having outer electronic configuration as,
\(\text{A} = 3s^2 3p^4, \text{B} = 3s^2 3p^5, \text{C} = 4s^2 4p^4, \text{D} = 4s^2 4p^5\)
identify the element with lowest ionization enthalpy \((\Delta_i \text{H}_1)\)