Match the following species with their respec | vedclass

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Question

(A) To find the number of electrons in each species:$(a)$ $O^{2-}$: Oxygen has atomic number $8$. $O^{2-}$ has $8 + 2 = 10$ electrons. Thus,$(a)-(iii)

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) To find the number of electrons in each species:$(a)$ $O^{2-}$: Oxygen has atomic number $8$. $O^{2-}$ has $8 + 2 = 10$ electrons. Thus,$(a)-(iii)$.$(b)$ $Li^{2+}$: Lithium has atomic number $3$. $Li^{2+}$ has $3 - 2 = 1$ electron. Thus,$(b)-(iv)$.$(c)$ $He$: Helium has atomic number $2$. It has $2$ electrons. Thus,$(c)-(ii)$.$(d)$ $Ca^{2+}$: Calcium has atomic number $20$. $Ca^{2+}$ has $20 - 2 = 18$ electrons. Thus,$(d)-(i)$.Therefore,the correct match is $a-iii, b-iv, c-ii, d-i$.

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 their respective number of electrons:
| Species | Number of electrons |
| :--- | :--- |
| $(a)$ $O^{2-}$ | $(i)$ $18$ |
| $(b)$ $Li^{2+}$ | (ii) $2$ |
| $(c)$ $He$ | (iii) $10$ |
| $(d)$ $Ca^{2+}$ | (iv) $1$ |

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

Species Number of Electrons

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

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

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

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

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Match the following species with their respective number of electrons:| Species | Number of electrons || :--- | :--- || $(a)$ $O^{2-}$ | $(i)$ $18$ || $(b)$ $Li^{2+}$ | (ii) $2$ || $(c)$ $He$ | (iii) $10$ || $(d)$ $Ca^{2+}$ | (iv) $1$ |