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Chemistry 36 Online
OpenStudy (anonymous):

Assign the position of the element having outer electronic configuration (i) ns2 np4 for n = 3 (ii) (n - 1)d2 ns2 for n = 4, and (iii) (n - 2) f7 (n - 1)d1 ns2 for n = 6, in the periodic table?

OpenStudy (anonymous):

@apoorvk @Callisto

OpenStudy (anonymous):

@UnkleRhaukus

OpenStudy (unklerhaukus):

(i) \(ns^2 np^4\) \(n=3\) \[\qquad 3s^23p^4\]

OpenStudy (unklerhaukus):

OpenStudy (anonymous):

so it belongs to 3 rd period and group 16

OpenStudy (unklerhaukus):

right,

OpenStudy (anonymous):

then (ii) (n - 1)d2 ns2 for n = 4

OpenStudy (anonymous):

it confuses

OpenStudy (unklerhaukus):

OpenStudy (anonymous):

it belongs to 4 th period but wat abt the group

OpenStudy (unklerhaukus):

\[(n-1)d^2\quad ns^2\qquad\qquad n=4\] \[3d^2\quad 4s^2\]

OpenStudy (anonymous):

ao group 4 or 14

OpenStudy (anonymous):

@UnkleRhaukus

OpenStudy (unklerhaukus):

?

OpenStudy (anonymous):

does this belongs to group 4

OpenStudy (anonymous):

or 14

OpenStudy (anonymous):

@UnkleRhaukus r u there!!

OpenStudy (unklerhaukus):

how could it be 14?

OpenStudy (anonymous):

yup u r correct it is 4

OpenStudy (unklerhaukus):

group 14 would be p block

OpenStudy (anonymous):

s

OpenStudy (anonymous):

then the last one

OpenStudy (anonymous):

6 th period and 3 rd group is it correct????? @UnkleRhaukus

OpenStudy (unklerhaukus):

(iii) \( (n - 2) f^7 (n - 1)d^1 ns^2\) for \(n = 6\) \[4 f^7 5d^1 6s^2\] \[\text{Gd}\]

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