n1 has 1
n2 has 4
If you add all the digits up, and the total equals a number that 3 can go in, then it's a factor of 3.
The mean, median and mode of one number MUST ALL BE that number.
Well 63 divided by 3 equals 21. So 3 times 21 equals 63. 21 is that number.
3 x 3 = 9
number added by 3 equals 62 = 59Let x = the numberEquation: x + 3 = 62x + 3 - 3 = 62 - 3x = 59
n=2 has 3 2p orbitals.
9
If the question is an attempt to ask "How many orbitals are there with principal quantum number n = 2", then 4 orbitals which can hold a total of 8 electrons.
principal energy level (n)= 3 Number of orbitals per level(n2)= 9 it equals 9 because it is n2 (32=9) n=1. 1 orbital n=2. 4 orbitals n=3. 9 orbitals n=4. 16 orbitals n=5. 25 orbitals n=6. 36 orbitalsn=7. 49 orbitals
n=3
5 sub-orbitals with (max.) two electrons in each, so 10 in total. This is also true for 4d and 5d orbitalsSymbols:dz2 , dxz ,dyz ,dxy ,dx2-y2
3
If you add all the digits up, and the total equals a number that 3 can go in, then it's a factor of 3.
9
There are three p orbitals in all levels 2 and above. these are the px, py and pz orbitals, the (suffix is the direction - px lies along the x axis). In the 5th level they will be 5px, 5py, 5pz
3
If it is in row 4 (4th period) it has 4 energy levels occupies. In the 2nd column (group 2A) it has 2 valence electrons. You ask how many orbitals it has. It has s and p orbitals. Is that what you mean? The configuration is 1s2 2s2 2p6 3s2 3p6 4s2 so there are 3 s orbitals and 6 p orbitals for a total of 9 orbitals.