Electronic configuration of ions of transition elements:
In first transition series atom loses 4s electron before they lose 3d electron.
the reason is that when electron goes in 3d orbital the energy of 3d orbital became less than 4s orbital .In other words 3d orbital became more stable than 4s.
once both the electron are lost from 4s orbital than electron are lost from 3d orbital.
for eg.
fe (26) [ Ar] 4s^2 3d^6
[fe^3+] [At] 3d^5 4s^0
Question
1 Write down electronic configuration of elements with atomic no. 24,46,74,57.
Ans.
electronic configuration of elements as:
1. cr(24) [Ar] 4s^1 3d^5
2. pd(46) [kr] 5s^0 4d^10
3. W(74) [xe] 4f^14 6s^2 5d^4
4. la(57) [xe ] 6s^2 5d^1
Question
2 cu^2+ is more stable than cu^+1 ?
Ans
cu^2+ cupric
cu^+ cuprous
cu^2+ compounds are more stable due to high lattice energy in solid state and high hydration energy in solution. and cu^2+size is small attain high energy as compare to cu^+1 .hence cu^2+ is more stable than cu^+1
In first transition series atom loses 4s electron before they lose 3d electron.
the reason is that when electron goes in 3d orbital the energy of 3d orbital became less than 4s orbital .In other words 3d orbital became more stable than 4s.
once both the electron are lost from 4s orbital than electron are lost from 3d orbital.
for eg.
fe (26) [ Ar] 4s^2 3d^6
[fe^3+] [At] 3d^5 4s^0
Question
1 Write down electronic configuration of elements with atomic no. 24,46,74,57.
Ans.
electronic configuration of elements as:
1. cr(24) [Ar] 4s^1 3d^5
2. pd(46) [kr] 5s^0 4d^10
3. W(74) [xe] 4f^14 6s^2 5d^4
4. la(57) [xe ] 6s^2 5d^1
Question
2 cu^2+ is more stable than cu^+1 ?
Ans
cu^2+ cupric
cu^+ cuprous
cu^2+ compounds are more stable due to high lattice energy in solid state and high hydration energy in solution. and cu^2+size is small attain high energy as compare to cu^+1 .hence cu^2+ is more stable than cu^+1
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