Home
Class 11
CHEMISTRY
The successive ionzation enegry values f...

The successive ionzation enegry values for an element `X` are given below:
(i) `1st` ionization enegry `= 410 kJ mol^(-1)`
(ii) `2nd` ionization energy `= 820 kJ mol^(-1)`
(iii) `3rd` ionization enegry `= 1100 kJ mol^(-1)`
(iv) `4th` ionization enegry `= 1500 kJ mol^(-1)`
(v) `5th` ionization enegry `= 3200 kJ mol^(-1)`
Find out the number of valence electrons for the atom `X`.

A

4

B

3

C

5

D

2

Text Solution

Verified by Experts

The correct Answer is:
A
The ionisation energies provide an indication about the number of valence electrons in an atom.
For example, Na has `IE_2 gt gt IE_1` because it has one
valence electron, while Mg has `IE_3 gt gt IE_2 gt IE_1`,
because it has two valence electrons.
In this case X has `IE_5 gtgt IE_4 gt IE_3 gt IE_2 gt IE_1`
Therefore it has four valence electrons.
Promotional Banner

Similar Questions

Explore conceptually related problems

The successive ionization energies for element X is given below IE_(1) : 250 kJ "mo"^(-1) IE_(2) : 820 kJ "mo"^(-1) IE_(3) : 1100 kJ "mo"^(-1) IE_(4) : 1400 kJ "mo"^(-1) Find out the number of valence electrons for the elements X.

Using the bond enthalpy data given below, calculate the enthalpy of formation of acetone (g). Bond enegry C-H = 413.4 kJ mol^(-1) , Bond enegry C-C = 347.0 kJ mol^(-1) , Bond enegry C=O = 728.0 kJ mol^(-1) , Bond enegry O=O = 495.0 kJ mol^(-1) , Bond enegry H-H = 435.8 kJ mol^(-1) , Delta_("sub")H^(Theta)C(s) = 718.4 kJ mol^(-1)

The first ionization energy of Na is 495.9 kJ mol^(-1) . What is the longest wavelength of light that could remove an electron from a Na-atom?

Delta_(f)H^(ɵ) of hypothetical MX is -150 kJ mol^(-1) and for MX_(2) is -600 kJ mol^(-1) . The enthalpy of disproportionation of MX is =-100 x kJ mol^(-1) . Find the value of x.