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The IP1, IP2, IP3, IP4 and IP5 of an element are 7.1, 14.3, 34.5, 46.8, 162.2 eV respectively. The element is likely to be:(a) Na(b) Si(c) F(d) Ca

Praneet Debnath , 8 Years ago
Grade 11
anser 3 Answers
Suraj Prasad

Last Activity: 8 Years ago

you can see that after removing the 4 electrons , the next Ionisation enthalpy is very high that signifies its now hard to take that out . Hence the element must be silicon out of these options given.

Umakant biswal

Last Activity: 8 Years ago

@ praneeth 
yes, option NO. B WILL TURN OUT TO BE THE CORRECT OPTION , AS WHEN U WILL SEE THE TREND IN IONISATION ENTHALPY , THEN in that case first it will increase upto the middle then become constant and then it will decrease .. 
but as per your given table its constantly increaseing but last 2 I.E are increaseing very rapidly . 
so. it will be obviously silicon as when u will remove the 2 valence or outer shell electron it will be make almost stable and stable config. will be prferrable by almost all element . 
HOPE IT CLEARS YOUR DOUBT 

 

Shankha Shubhra Giri

Last Activity: 7 Years ago

46.8 to 162.2 is much greater change in IP value .So for IP value 4 it will have 4 valence electrons and silicon has 4 valence electrons. Therefore answer is (b)

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