why does chromium (element 24) has 3d^5 and 4s^1 instead of 3d^4 4s^2?

Question

CAN SOME ONE HELP ME ELECTRON CONFIGURATION

why does chromium (element 24) has 3d^5 and 4s^1 instead of 3d^4 4s^2

Answer 1

By Hund's rule of multiplicity, the more or most stable state of electronic configuration is such that all the electrons are at most UNPAIRED before Pairing up.

In Chromium, there is an option for the electrons to "scatter" in five separate unpaired d orbitals first before being "forced" to pair up. Moreover, the electron in the 4s level is unpaired as well. The stabilizing effect of these unpaired spins makes up for a lot more than the stabilizing effect of the paired 4s2 electrons.

Answer 2

There are several parameters that determine the energy of an electron orbital and they don't all scale the same way with the number of electrons or atomic number Z. The first quantum parameter, n, makes a big difference at low Z but its importance fades quickly as Z increases. The jump in energy from 1s to 2s is larger than from 2s to 3s. The 3s-4s difference is even smaller, and so on. Other quantum parameters, like L, ML, may be less sensitive to atomic number, and as the influence of N shrinks, L may become the dominant term, dictating that 4s is overall cheaper than 3d.

That's not exactly the quantum-mechanical explanation, but that's as well as I understand it even after reading the Wikipedia article.

[Edit] The Pauli exclusion principle that leads to Hunds Rule might have an energy associated with it, and this energy may become dominant when n=3 or 4, overruling the effect of n.

Answer 3

It belongs to D-block elements. We can represent the configuration of these elements by (n-1)d1-10ns0,1,2.
Chromiun belongs to 4th period . It has ten elements Sc-Zn see the periodic table. Filling of unoccupied orbitals (3d) takesplace. Occasionally an electron from 4s orbital is shifted out of turn to 3d orbitaldue to higher stability of half filled or completely filled orbitals. Eg- Cr, Cu. Have configration- 3d5,4s1 and 3d10,4s1 instead of expected 3d4,4s2 and 3d9, 4s2. After the 3d level is filled in the next six elements of this period the 4p orbitals are gradually filled and Kr had the config.(Ar) 3d10,4s2,4p6.

Answer 4

A half filled orbitial is more stable (in this case). Pd normally just fills its 4d orbitial and has none in the 5s. The transition metals are more complicated than most of the other elements as far as electron configuration. By way of example Mo has 5 different oxydation states (+2,3,4,5,6).

[edit] Oh yeah, Hund's Rule. Electrons prefer to be unpaired because like charges repel eachother.

Answer 5

When you calculate the energy of the electrons you find that the rule first fill the shells with lowest number first the the subshell with lowest number does not hod for shells m, n &. Remember energy of binding . The absolute value of that energy is BZ^2/ n ^2 , where n is the number of the shell. When n=3,4 & , the diffrences of energy binding are so low that the rule of filling holds no more and in order to explain the filling of shells, you must caculate the energies for each electron

Answer 6

Let's see if I can simplify a very complicated subject. No quantum physics I assure you. Get your Periodic Table and follow along as you read this. Imagine each shell of the atom was a hotel with different numbers of rooms. As you travel out from the nucleus the hotels have more rooms but generally the rooms are more expensive (lower energy levels actually but we're still pretending they are hotel rooms). Since the rooms nearer the nucleus are cheaper they naturally fill up first. OK, here goes. The first hotel has only two rooms. H fills one and He fills the other. First Period. Down the street is the next hotel with eight slightly more expensive rooms - Li, Be, B, C, N, O , F and Ne fill these rooms. This is period two. Now, things get a little more complicated. Hotel number three has eighteen rooms but eight of them are only slightly more expensive than those in hotel number two. The remaining ten are considerably more expensive. So Na thru Ar, eight elements (sorry, hotel guests) fill these rooms. This is period three. Now, this is where it gets tricky. Remember I said there was a gap in price between the first eight rooms and the remaining ten rooms in hotel three? Well, the owners of hotel three and four are very competitive. The owner of hotel four further down the street prices two of his rooms cheaper than the ten rooms in hotel three. So naturally K and Ca fill those rooms. This is the beginning of period four. Now, the rest of hotel four's rooms are very expensive so ten elements (darn, I forgot, hotel guests), Sc through Zn fill the remaining ten rooms in hotel three. These are the transition elements of period four. Almost there. Remember I said hotel owners three and four were very competitive and not above stealing business from each other? Well, hotel owner three told his manager he would give him a bonus when those ten rooms were either half full or full. So when the fourth element arrives (Cr) the manager goes down the street and gets an electron out of hotel number four. Remember hotel four had only two electrons at this point. Now it has one. That's why the fourth shell of chromium has only one electron and the third shell has one more than you would expect. When Manganese enters the picture its electron goes down the street restoring shell four to its occupancy of two while shell three remains half full and the manager gets his bonus. Then we have Fe, Co and Ni. Then with copper the same thing happens. Hotel three is one away from being full so in order to collect his bonus early the manager again steals an electron from hotel four leaving only one electron in hotel four and one more than you would expect in hotel three. Then with zinc, balance is restored. Hotel three is full, there are two electrons in hotel four and the remaining six elements, Ga through Kr go into hotel four, joining the two who are already there to make a total of eight (an octet). At this point period four is complete. I know this is complicated but study this over and over following along on your Periodic Table until it becomes clear. Then you can forget about hotel rooms and think about electronic structure. But remember, it's simply a matter of increasing energy levels (price of rooms). This is a silly example and below the level of some purists but it has always helped my students understand a very complicated subject. BTW. I typed this from memory at four in the morning so please forgive me if I have made any minor errors.

Answer 7

coz half filled orbitals are more stable than incomplete orbitals and 3d^5 and 4s^1 shows that both 3d and 4s are half filled. whereas in 3d^4 4s^2, 3d orbital has an unstable configuration

Answer 8

A d subshell that's 0.5-crammed or complete (ie 5 or 10 electrons) is greater stable than the s subshell of the subsequent shell. it is the case simply by fact it takes much less potential to maintain an electron in a nil.5-crammed d subshell than a crammed s subshell.