An ionic compound A^{+} B^{-} is most likely | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The ionization energy $(IE)$ is the energy required to remove an electron from a neutral gaseous atom to form a cation. Lower $IE$ facilitates the

Vedclass · Accepted Answer

the ionization energy of $A$ is low and electron affinity of $B$ is high

Vedclass · Answer

(B) The ionization energy $(IE)$ is the energy required to remove an electron from a neutral gaseous atom to form a cation. Lower $IE$ facilitates the formation of $A^{+}$. Electron affinity $(EA)$ is the energy released when an electron is added to a neutral gaseous atom to form an anion. Higher $EA$ facilitates the formation of $B^{-}$. Therefore,for the formation of an ionic compound $A^{+} B^{-}$,the ionization energy of $A$ should be low and the electron affinity of $B$ should be high.

An ionic compound $A^{+} B^{-}$ is most likely to be formed when

Similar Questions

$3$ and $6$ electrons are present in the outermost orbit of $A$ and $B$ respectively. The chemical formula of its compound will be

Which of the following compounds has the highest lattice enthalpy?

Given below are two statements:
Assertion $(A)$: Ionic compounds are formed by non-directional bonds.
Reasoning $(R)$: They are soluble in nonpolar solvents.
The correct answer is:

Which of the following statements is correct for $CsBr_3$?

Which among the following compounds shows the highest lattice energy?

Vedclass Test Series

Exam Paper Generator

Online Exam Module