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Question

(D) The number of electrons in a species is calculated by subtracting the charge from the atomic number of the neutral atom.$1$. For $Be^{2+}$: Atomic

Vedclass · Accepted Answer

$(i-c), (ii-a), (iii-b), (iv-d)$

Vedclass · Answer

(D) The number of electrons in a species is calculated by subtracting the charge from the atomic number of the neutral atom.$1$. For $Be^{2+}$: Atomic number of $Be$ is $4$. Electrons = $4 - 2 = 2$. Thus,$(i-c)$.$2$. For $H^{+}$: Atomic number of $H$ is $1$. Electrons = $1 - 1 = 0$. Thus,$(ii-a)$.$3$. For $Na^{+}$: Atomic number of $Na$ is $11$. Electrons = $11 - 1 = 10$. Thus,$(iii-b)$.$4$. For $Mg^{+}$: Atomic number of $Mg$ is $12$. Electrons = $12 - 1 = 11$. Thus,$(iv-d)$.Therefore,the correct match is $(i-c), (ii-a), (iii-b), (iv-d)$.

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

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

Species Number of Electrons

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

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

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

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

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Match the following species with the correct number of electrons present in them: Species Number of Electrons $(i) Be^{2+}$ $(a) 0$ $(ii) H^{+}$ $(b) 10$ $(iii) Na^{+}$ $(c) 2$ $(iv) Mg^{+}$ $(d) 11$