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info@tutorhelpdesk.com About Chemistry: Chemistry is another physical science which has its roots attached to Physics. It is a study of the composition and the behaviours of matter. The basic unit of the study of chemistry is considered to ‘atom.’ Etymology of the word chemistry can be traced to the word ‘alchemy’ which has been practiced largely in the Middle Eastern Countries of the world. The term alchemy on the other hand traces its roots to the Persian word kimia. Many scholars have termed Chemistry as a core science as it connects the relation between other fields of sciences such as biology, physics, geology etc. Jean-Baptiste Dumas had defined the word "chemistry" as referring to the science concerned with the laws and effects of molecular forces. This definition was further evolved to mean the science of substances: study of their structure, properties, and the reactions that change them into other substances. The study of chemistry assignment help can be traced to the widely prevalent process of burning that led to the study of metallurgy. The desire for gold led to the discovery of the process for its purification. The findings further replicated and used in the experiments for chemical substances by burning the substances under certain temperature.

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Q 1. Explain why orbitals like 1p, 2d and 3f are not possible?
Solution:
This is because the sets of quantum numbers required for these orbitals are not allowed.
1p refers to n = 1, l = 1 which is not allowed because values of I are governed by n. That is, l = 0, 1, 2..., (n-1). Thus, for n = 1; l = 0 and l cannot have a value equal to one.
Similarly: 2d refers to n = 2, l = 2 which again is not possible. For n = 2; l can have values 0, 1 only.
That is l = 2 is not permissible in this case.
Also 3f is not possible because for 3f, the combination n = 3, l = 3 is again not allowed. For n = 3, l can’t have a value higher that (n-1), that is, 2. In other words, for n =3; permissible l = 0, 1 and 2 only.
Q 2. Which sets of quantum numbers are not permissible and why?
Solution:
(i) This is allowed. Because for n = 5; l = 0, 1, 2, 3, 4; m = 0, ± 1, ± 2, ± 3, ± 4; s = ± ½. The values in question are permissible.
(ii) It is not allowed because value of m (+l,...., 0,....,-l) are governed by l and when l = 0, m cannot have values other than zero (m=0).
(iii) It is not allowed because when l = 1, m is allowed values = = 1, 0, - 1 only. A value m = 2 is not permissible.
(iv) It is not allowed because values of l are governed by n. For n = 2; l can have values equal = 0, 1 only. Therefor, n =2, l = 2, is not permissible.

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Q 3. Calculate and identify all the individual orbitals for (i) = 3 (ii) n = 4
Solution:
(i) n= 3; l = 0 1 2; m =0, ± 1, ± 2 :. Total no. Of orbitals = 1 + 3 + 5 = 9.
These are: 3s; 3px’ 3py; 3pz; 3dxy’ 3dyz’ 3dxz’ 3dz2, 3dx2-y2 as detailed in Table 1.1.
(ii) For n = 4; l = 0, 1, 2, 3; m = 0, ± 1, ± 2, ± 3. Individual orbitals are:
L m orbitals
0 0 4s
1 ± 1, 0 4px’ 4py’ 4pz
2 ± 2, ± 1, 0 4dxy’ 4dyz’ 4dxz’ 4dz2’ 4dx2-y2
3 ± 3, ± 2, ± 1, 0 4fx3’ 4fy3’ 4fz3’ 4fxyz’ 4f(x2-y2)’
4f(x2-y2)’ 4f(x2-y2)’
For n = 4: total no. Of orbitals = 1 + 3 + 5 + 7 = 16.
Q 4. Which of the following subshells are permissible and why 5g, 6h, 3d?
Solution: 5g is permissible. Because for n = 5; l can have values l = 0, 1, 2, 3, 4 with their subshells 5s, 5p, 5d, 5f, 5g, respectively.
6h is also permissible. Because for n = 6; l can have values l = 0, 1, 2, 3, 4 which correspond respectively to 6s, 6p, 6d, 6f, 6g, 6h.
3d is also permissible. Because for n = 3; l can have values l = 0, 1, 2 which correspond respectively to 3s, 3p, 3d.
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