PLE give correct explanation of my question 24
Answers
Answer:
a) element is sulphur
Period =3,group =6
b)element is titanium
Period =4,group =4—b
POINTS TO REMEMBER
• Period always equal to the principal quantum no. ie, n .
• Group
(i) if s-block , then no. of valence electrons
(ii) if p-block , then no. of electrons in the valence shell (no. of electrons in p orbital + no. of electrons in the s orbital) + 10
(iii) if d-block , then no. of electrons in the valence shell (sum of no. of electrons in (n-1) th d orbital and the n th p and s orbitals)
(iv) if f-block , then group is always equal to 3
Atomic number : Write out the electronic configuration and count the number of electrons. (Its always better to learn the atomic number of the noble gases and just write out the rest for the other elements. For example if you want to write the Atomic no. of the elements in question (c) here , you know it comes in the 6 th period , so write the Atomic no. of the noble gas of 5 th period then write the rest of the configuration ie, [Xe] 4f7 5d1 6s2 . Atomic no of Xe + 7 + 1 + 2 = 54 + 10 = 64 )
ANSWERS
(a) 3s2 3p4
Period = 3(n)
Group = 2(s) + 4(p) = 6 = 16th
Block = p-block
Element = Sulphur (S)
Atomic no. = 16
(b) 3d2 4s2
Period = 4(n)
Group = 2(d) + 2(s) = 4th
Block = d-block
Element = Titanium (Ti)
Atomic no. = 22
(c) 4f6 5d1 6s2
Period = 6 (n)
Group = 3 (All f-block elements belong to group 3)
Block = f-block
Element = Gadolinium (Gd)
Atomic no. = 64 (found above in the ex0lanation part)
Most probably the name of the element won't be asked since many elements (for example the element in the question (c) ) are not learnt. Students are advised to learn only the first 4 periods , though studying the whole table is very good.
Anyway, Happy 150 th Anniversary of the Periodic Table !!
And ...
Hope this helps
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