Question

The equilibrium internuclear distance is 246 pm for an ionic compound A-X. Calculate the
dipole moment of the ionic compound in gas phase.
F
ST
(A) 3.94x10-Cm
(B) 3.94x10- Cm
(C) – 3.94x10*2 Cm
(D) 3.94x10° Cm
TRE
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Answer 1

Answer: 3.94 * 10⁻²⁹ C-m

Step-by-step explanation:

The equilibrium inter-nuclear distance for an ionic compound i.e., the bond length, r = 246 pm

1 pm = 10⁻¹² m

∴ r = 246 * 10⁻¹² m

Charge of 1 e⁻, Q  = 1.6 * 10⁻¹⁹ Coloumb

We know,  

The formula for the dipole moment (µ) is given as,

µ = Q * r

Substituting the given values

⇒ µ = (1.6 * 10⁻¹⁹ C) * (246*10⁻¹² m)

⇒ µ = 393.6 * 10⁻³¹ C-m

⇒ µ = 3.936 * (10⁻³¹ * 10²) C-m

⇒ µ = 3.94 * 10⁻²⁹ C-m

Thus, the dipole moment of the ionic compound A-X in the gaseous phase is 3.94 * 10⁻²⁹ C-m.