
The hydride ion is isoelectronic with:
A. ${H^ + }$
B. $H{e^ + }$
C. $He$
D. $Be$
Answer
559.2k+ views
Hint: We need to know what are isoelectronic species and accordingly analyse the isoelectric nature of hydride ions. Atoms and ions which have the same number of electrons are said to be isoelectronic atoms or ions. In other words, species with the same electronic configuration are isoelectronic. Example of isoelectronic species are ${N^{3 - }}$ , ${O^{2 - }}$ whose electronic configuration is \[1{s^2}2{s^2}2{p^6}\] . Now we can analyse the hydride ion.
Complete step by step answer:
We know that isoelectronic species have the same number of electrons that have the same electronic configuration. Given is a hydride ion whose molecular formula is $H^-$. The negative charge says that it has an extra electron since the electron has a negative charge. In general, hydrogen H has one electron. Adding an extra electron changes it to hydride ions. Hence hydride ions have a total of two electrons.
Now we calculate the number of electrons in the given options:
$H^+$: The positive charge indicates that an electron has been lost from the hydrogen. Hence it has no electrons.
$He^+$: The positive charge indicates that an electron has been lost from the helium atom. Helium has an atomic number 2 and hence has 2 electrons. The loss of one electron leaves it with one electron.
He: Helium has an atomic number 2 and hence has 2 electrons. Since there is no charge on the atom, there is no loss or gain of electrons which is isoelectronic with the hydride ion.
Be: Beryllium has an atomic number 4 and hence has 4 electrons. Since there is no charge on the atom, there is no loss or gain of electrons.
So, the correct answer is Option C.
Note: It must be noted that hydride ion is also known as hydrogen anion. The electronic configuration of hydrogen anion is $1{s^2}$ and that of Helium is also $1{s^2}$ . Hence it is isoelectronic. Hydride ion can also be isoelectronic with various other species such as $L{i^ + }$ , $B{e^{2 + }}$ etc. For atoms or ions that are isoelectronic, the number of protons determines the size.
Complete step by step answer:
We know that isoelectronic species have the same number of electrons that have the same electronic configuration. Given is a hydride ion whose molecular formula is $H^-$. The negative charge says that it has an extra electron since the electron has a negative charge. In general, hydrogen H has one electron. Adding an extra electron changes it to hydride ions. Hence hydride ions have a total of two electrons.
Now we calculate the number of electrons in the given options:
$H^+$: The positive charge indicates that an electron has been lost from the hydrogen. Hence it has no electrons.
$He^+$: The positive charge indicates that an electron has been lost from the helium atom. Helium has an atomic number 2 and hence has 2 electrons. The loss of one electron leaves it with one electron.
He: Helium has an atomic number 2 and hence has 2 electrons. Since there is no charge on the atom, there is no loss or gain of electrons which is isoelectronic with the hydride ion.
Be: Beryllium has an atomic number 4 and hence has 4 electrons. Since there is no charge on the atom, there is no loss or gain of electrons.
So, the correct answer is Option C.
Note: It must be noted that hydride ion is also known as hydrogen anion. The electronic configuration of hydrogen anion is $1{s^2}$ and that of Helium is also $1{s^2}$ . Hence it is isoelectronic. Hydride ion can also be isoelectronic with various other species such as $L{i^ + }$ , $B{e^{2 + }}$ etc. For atoms or ions that are isoelectronic, the number of protons determines the size.
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