Question

Copper forms two chlorides CuCl and CuCl2 (1) Identify the copper ions in these compounds? (2) What is the valency of copper in each of these compounds? (3) By writing the sub-shell electronic configuration of copper, find out a sub-shells from which the electrons are lost to form these ions. (4) Write the period and group to which copper belongs. (Hint: Atomic number of copper =29)

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Answer 1

SOLUTION :-

Given,

Copper forms two chlorides CuCl and

Atomic number of copper (Cu) = 29.CuCl2

(1) The copper ions in the CuCl and CuCl2 compounds are

CuCl − Cu+

CuCl2− Cu2+

(2) The valance of copper in

CuCl is 1

CuCl2 is 2.

(3) Electronic configuration

Cu (29)− 1s2 2s2 2p6 3s2 3p6 3d10 4s1.

In the case of CuCl electron is lost from 4s orbital.

In the case of CuCl2 electron is lost from 4s orbital and another electron is lost from 3d orbital.

The electronic configuration of Cu2+ is 1s2 2s2 2p6 3s2 3p6 3d9.

(4) Copper (Cu) belongs to 4th period and 11th group in the periodic table.

Answer 2

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Given:-

Copper forms two chlorides CuCl and

$</p><p></p><p> {\rm{Atomic \: number \: of \: copper (Cu) =29CuCl_2}}</p><p></p><p>$

${\rm{(1) \: The \: copper \: ions \: in \: the CuCl\: and}}$${\rm{CuCl_2\: compounds\: are}}$

${\rm{CuCl -{ Cu}^{ + }}} \ \\ {\rm{ CuCl_2 -{ Cu}^{2 + }}}$

${ \rm{(2) \: The \: valance \: of \: copper \: in}}$

${\rm{CuCl \: is \: 1}} \\ {\rm{{CuCl_2 \: is \: 2}}}$

${\rm{(3) \: Electronic \: configuration}}$

${\rm{Cu(29) - \: 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{1}}}$

In the case of CuCl electron is lost from 4s orbital.

${\rm{In \: the \: case \: of \: CuCl_2 electron \: is}}$

lost from 4s orbital and another electron is lost from 3d orbital.

${\rm{The \: electronic \: configuration \: of}}$

${\rm{Cu^{2 + } is \: 1{s}^{2}2{s}^{2}{2p}^{3}{3s}^{2} {3p}^{6} {3d}^{9}}}$

(4) Copper (Cu) belongs to 4th period and 11th group in the periodic table.

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