Match the following species with the correct number of electrons present in them:

Species Number of Electrons

$(i) {Be}^{2 +}$ $(a) 0$

$(ii) H^{+}$ $(b) 10$

$(iii) {Na}^{+}$ $(c) 2$

$(iv) {Mg}^{+}$ $(d) 11$

$(e) 4$

Question

Match the following species with the correct number of electrons present in them:

Species	Number of Electrons
$(i) {Be}^{2 +}$	$(a) 0$
$(ii) H^{+}$	$(b) 10$
$(iii) {Na}^{+}$	$(c) 2$
$(iv) {Mg}^{+}$	$(d) 11$
	$(e) 4$

Aicharya · Accepted Answer

We know that the number of electrons is equal to the atomic number of element in neutral state.

Atomic number of beryllium is $4$ but ${Be}^{2 +}$ can be formed by losing the two electrons so remaining electrons in ${Be}^{2 +}$ will be $4 - 2 = 2$ .

Similarly, the number of electrons in $H^{+}$ is $1 - 1 = 0$ .

Similarly, the number of electrons in ${Na}^{+}$ is $11 - 1 = 10$ .

Similarly, the number of electrons in ${Mg}^{+}$ is $12 - 1 = 11$ .

Hence, the correct match will be $(i - c), (ii - a), (iii - b), (iv - d)$ .

Match the following species with the correct number of electrons present in them: Species Number of…

Solution