1Look , this is the general approach.
Given any two compounds , the one with lower IE is more metallic,rihgt ?
i.e. it can easily lose the electrons. which is implied by the amt of energy required to remove the electrons.
What I've said above is for differentiating which of the given elements has more metallic character.
If you're given only one element and asked to given the number of valence electrons or similarly the highest ON or most stable ON. or something like that , we proceed like this
SEE, W.K.T IE 1 < IE2 < IE3 and so on.
this is true for any element , right?
NOW , REMOVAL OF ELECTRONS FROM OCTET CONFIG OR WHEN ELEMENT HAS LIBERATED A GOOD AMT OF EXCHANGE ENERGY IS DIFFICULT.( STABLE MOLECULES I MEAN TO SAY DUE OCTET OR PRESENCE OF UNPAIRED e-s IN ORBITALS OF SIMILAR ENERGY. ) EX : NOBLE GASES - OCTET . NITROGEN - HAS HIGHER IE1 THAN OXYGEN BECAUSE OF EXCHANGE ENERGY .
WHENEVER THERE IS A WHOOPING INCREASE IN THE AMT OF ENERGY REQUIRED TO REMOVE ELECTRON , THE ATOM HAS ONE OF THESE ABOVE MENTIONED STATES . I.E. STABLE STATES. THE VALENCE E-S ARE EASY TO REMOVE SO LESS ENERGY IS REQUIRED , WHENEVER STABLE STATE IS ATTAINED BY ATOM , FURTHER REMOVAL BECOMES DIFFICULT as in the uestion . The atom is a trivalent metal so , less E is required to remove the first 3 e-s . after that it attains octet , so E required to remove the electron increases dratically due to many factors. First , you'er removing the e from the triply +vely charged atom and also you're disrupting octet. hope you've understood.
If you've not , then ,
I'VE GOT ONE QUESTION , why is the e.c. of Cr [Ar] 4s1 3d5 and not [Ar]4s2 3d4 similarly with Cu . If you can explain the reason ,then you can definitely understand.