Answer: 1s2 2s2 2p3
Explanation:
atomic number is 7 hence it shuld be 2,5 if numerically.. tho the standard notation is the above giver answer
Answer:
b. The splitting of the d-orbitals is smaller in the [Ni(Cl)6]4- complex than in the [Ni(en)3]2+ complex.
Explanation:
The spectrochemical series is an arrangement of ligands in increasing order of their magnitude of crystal field splitting.
Ligands that occurs towards the right in the series are called strong field ligands and they tend to cause a greater magnitude of crystal field splitting. Ligands that occur towards the left hand side in the series are called weak field ligands and they tend to cause a lesser magnitude of crystal field splitting.
Since Cl^- is a weak field ligand, it causes a lesser magnitude of d orbital splitting compared to ethylenediammine (en) which causes a greater magnitude of d orbital splitting.
Hence; the splitting of the d-orbitals is smaller in the [Ni(Cl)6]4- complex than in the [Ni(en)3]2+ complex.
Answer:
Concentration of nitrate in the new solution = 0.007 M
Explanation:
Given:
Concentration nitrate solution = 0.070 m
Volume of aliquote of the nitrate solution is add = 10.0 ml
Total volume = 100 ml
Find:
Concentration of nitrate in the new solution
Computation:
Number of M. mole = 0.070 m x 10.0 ml
Number of M. mole = 0.7 m-moles
Concentration of nitrate in the new solution = 0.7 m-moles / 100 ml
Concentration of nitrate in the new solution = 0.007 M
Answer:
0
Explanation:
There are no unpaired electrons in the given element. It must be noted that for the atom above, we have even numbered electrons. The total electron we are having here is 18.
Now, we must also know that while the s orbital is not degenerate, the P orbital is degenerate. What this mean is that the p orbital is broken down into three different sub orbitals which is the Px , Py and Pz. Hence we can see that there are 6 electrons to enter into the P orbital too.
We can see that all the S orbitals have been completely filled with two electrons alike each. This is also the case for the P orbital as the 3 suborbitals take in 2 each to give a total of six
