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# The charge on cobalt in ${[Co{(CN)_6}]^{3 - }}$ is:A. $- 6$B. $- 3$C. $+ 3$D. $+ 6$

Last updated date: 14th Sep 2024
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Hint: In the compound the cyanide group is present which has charge equal to $- 1$. And we know that the transition elements can show more than one type of valency so here we have to use charge balance to calculate the charge on cobalt.

Transition elements: Those elements which are in groups from three to eleven. They are called a transition because they are in between the s- block elements and p-block elements. For example: Scandium, iron, zinc, etc. They have fully or at least one electron in their d-orbits. For example: scandium has one d-electron, zinc has ten d-electrons. Due to the different number of electrons present in their d-shells they show different valencies. For example: Scandium having one electron can show three valencies as one, two and three. Because the atomic number of scandium is $21$and its electronic configuration is $Ar3{d^1}4{s^2}$. By losing only one s-electron it will attain one valency and by losing both the s-electrons it will attain valency two. And by losing two s-electrons and one d-electron it will achieve three valencies which is a stable state of scandium. Because after losing three electrons it will attain electronic configuration of noble gas argon.
In the question we are given the compound of cobalt. We know that the atomic number of cobalt is $27$. So its electronic configuration will be $Ar3{d^7}4{s^2}$. So it can show a number of valencies by losing seven d-electrons and two s-electrons. In the compound given in the question cyanide group is attached to the cobalt. And the charge on the cyanide group is $- 1$. So the total charge due to the cyanide groups will be $- 6$. Now the whole charge of the compound is given as $- 3$. And let the charge on cobalt be $x$. Then the total charge due to the cobalt and cyanide groups must be equal to $- 3$.
$\Rightarrow \ x - 6 = - 3 \\ \Rightarrow\ x = 3 \\$