
Which form of silver is colourless?
Answer
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Hint :For a transition metal present in the d block elements if you need to find if the metal is coloured or not or in which state it is colourless, if the electrons in the outermost d and s orbitals are paired then colourless and unpaired then some colour will be exhibited.
Complete Step By Step Answer:
Silver is present in the d block and is in the eleventh period of the periodic table. It has 47 electrons in its outermost shell. It exhibits the highest electrical conductivity, thermal conductivity and reflectivity of any metal. It is a soft, white, lustrous metal.
The electronic configuration of silver is $[Ar]4{d^{10}}5{s^1}$using the rules of giving the electronic configuration. The s orbital isn’t filled and d orbital is filled because half filled or full filled orbitals are more stable than any other configuration.
From the configuration given above we can say that the electrons are filled in the d orbital and the s orbital has only one unpaired electron.
Now for a transition metal to exhibit any colour they must have incompletely filled $(n - 1)d$orbitals. This is asked because in the case of incompletely filled orbitals, when the light photons interact with unpaired electrons, they emit colour.
Now for silver to be colourless it needs to have no unpaired electron.
If we lose one electron in the outermost orbital, then
$A{g^ + } = \,[Ar]5{d^{10}}4{s^0}$
Now silver in this excited state has no unpaired electron and this form of silver will be colourless.
$A{g^ + }$is the form of silver which will always be colourless.
Note :
Always remember the filling up of orbitals in the outermost orbitals is to be done such that we get a stable structure and that exists when the orbitals are completely filled or partially filled.
Complete Step By Step Answer:
Silver is present in the d block and is in the eleventh period of the periodic table. It has 47 electrons in its outermost shell. It exhibits the highest electrical conductivity, thermal conductivity and reflectivity of any metal. It is a soft, white, lustrous metal.
The electronic configuration of silver is $[Ar]4{d^{10}}5{s^1}$using the rules of giving the electronic configuration. The s orbital isn’t filled and d orbital is filled because half filled or full filled orbitals are more stable than any other configuration.
From the configuration given above we can say that the electrons are filled in the d orbital and the s orbital has only one unpaired electron.
Now for a transition metal to exhibit any colour they must have incompletely filled $(n - 1)d$orbitals. This is asked because in the case of incompletely filled orbitals, when the light photons interact with unpaired electrons, they emit colour.
Now for silver to be colourless it needs to have no unpaired electron.
If we lose one electron in the outermost orbital, then
$A{g^ + } = \,[Ar]5{d^{10}}4{s^0}$
Now silver in this excited state has no unpaired electron and this form of silver will be colourless.
$A{g^ + }$is the form of silver which will always be colourless.
Note :
Always remember the filling up of orbitals in the outermost orbitals is to be done such that we get a stable structure and that exists when the orbitals are completely filled or partially filled.
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