# General Knowledge - Chemistry - Discussion

Discussion Forum : Chemistry - Section 1 (Q.No. 5)
5.
The number of d-electrons in Fe2+ (Z = 26) is not equal to that of
p-electrons in Ne(Z = 10)
s-electrons in Mg(Z = 12)
d-electrons in Fe(Z = 26)
p-electrons in CI(Z = 17)
Explanation:
No answer description is available. Let's discuss.
Discussion:
31 comments Page 1 of 4.

Clevein said:   10 months ago
The electron configurations are as follows:

1. **Fe²⁺ (Z = 26):** [Ar] 3d⁶.
=> The number of d-electrons is 6.

2. **Ne (Z = 10):** 1s² 2s² 2p⁶
=> There are no p-electrons in Ne.

3. **Mg (Z = 12):** 1s² 2s² 2p⁶ 3s²
=> The number of s-electrons is 2.

4. **Fe (Z = 26):** [Ar] 4s² 3d⁶
=> The number of d-electrons is 6.

5. **Cl (Z = 17):** 1s² 2s² 2p⁶ 3s² 3p⁵
=> The number of p-electrons is 5.

In summary:

- The number of d-electrons in Fe²⁺ (Z = 26) is **6**.
- The number of p-electrons in Ne (Z = 10) is **0**.
- The number of s-electrons in Mg (Z = 12) is **2**.
- The number of d-electrons in Fe (Z = 26) is **6**.
- The number of p-electrons in Cl (Z = 17) is **5**.

Therefore, the statement is correct.

The number of d-electrons in Fe²⁺ is not equal to the specified electron counts in Ne, Mg, Fe, and Cl.

Ramanuj said:   9 years ago
To start up, the electron configuration of Fe atom is 1s2 2s2 2p6 3s2 3p6 4s2 3d6.

Since the problem is Fe2+ ion, than two electrons are removed from the atom so the correct configuration is 1s2 2s2 2p6 3s2 3p6 3d6.

Likewise, re-check your second configuration. I'm assuming you intended to write it as 1s2 2s2 2p6 3s2 3p6 4s2 3d4. This is incorrect because by then, the element would not be Fe2+ ion. Instead, this would be the electronic configuration of "chromium atom".

Basically, 2 electrons in the 4s orbital are removed because it is much farther away from the nucleus than 3d orbital. Since the electrons in 4s are farther and have weaker pull from the nucleus, there's more tendency for them to be dislodged from the atom.
(1)

Osama Shata said:   2 years ago
Fe (Z=26) configuration is [Ar]18 4S2,3d6.

Fe2+ means it will lose two electrons from its outer shell which is 4S.

So the configuration is [Ar]18 4s0, 3d6 . here the Fe2+ is stable because orbitals are fully filled or fully empty. d-electrons Numbers are 6.

Ne10 configuration is 1S2,2S2,2p6. p-electrons Numbers are 6
Mg12 configuration is 4s2,2s2,2p6,3s2 so s-electrons Numbers are 6
Cl17 configuration is 4s2,2s2,2p6,3s2,4p5. After applying half filled orbitals rule, the configuration will be 4s2,2s2,2p6,3s1,3p6 so p-electrons Numbers are 12
So, D is the correct answer.
(3)

In case of Fe(Z=26):[Ar]18,3d6,4s2. here is 6 ele.in d-orbital.
Fe+2 two ele. will be lose, so Fe+2:[Ar]18,3d5,4s1.one ele is lost from 4s and one ele. is from 3d orbital it is bcoz if the d-orbital is half filled or fully filled atom will gain more stability.so here is 5 ele in d orbital.
Cl(Z=17): 1s2,2s2,2p6,3s2,3p5.here 5 ele. in p-orbital.
Mg(Z=12): 1s2,2s2,2p6,3s2.here 2 ele.in s-orbital.

Shikamaru Nara said:   1 decade ago
The 4s orbital is at a higher energy level than the 3d orbitals and so because it is Fe(2+) the two electrons both leave the 4s orbital, leaving the 3d orbital with 6 electrons. And so:

No.of p electrons in Ne is 6(2p6).

No.of s electrons in Mg is 6(1s2,2s2,3s2).

No.of d electrons in Fe is 6 (d6).

No.of p electrons in Ci is 11(2p6,3p5).
(1)

Because if we write firstly the configuration of Fe we get 6 electrons in d-orbital of Fe, but in the question Fe2+is given so we should take of two electrons in the d-orbital of Fe, then we will remain with only 4electrons in d-orbital of Fe. Now if we see options cl will get 4electons in p-orbital.

In this case, Fe(26)is 1s2,2s2,2p6,3s2,3p6,3d6,4s2...
in Fe2+ is 1s2,2s2,2p6,3s2,3p6,3d5,4s1...because of half-filled configuration is stable... in Ne,Mg,Fe these are elements having full filled configuration and stable configuration.... Cl only having Ne(10) 3s2,3p5..its a unstable configuration...

Sarawar said:   9 years ago
Actually half filled and full filled orbital are much more stable than partially filled orbital, that is why fe will first loose one electron from d orbital to half filled the 3d orbital then loose one other electron from 4s orbital and become more stable.