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The right answer is A. Boron has three valence electrons in his outer shell and there are 4 electrons around him in BF4-. So 3-4=-1 +3 is the OXIDATION NUMBER which is differently from formal charge.
Formal charge is defined as the difference between the number of valence electrons and the number of Lewis electrons,i.e all unshared electrons on the atom plus the half of all the electrons shared by the atom in the Lewis structure.
Oxidation number is defined as the difference between the number of valence electrons and the number of electrons the atom has in the compound. The number of electrons the atom has assigned in the compound includes all unshared electrons plus half of the electrons shared with atoms of equal electronegativity plus all electrons shared with atoms of lesser electronegativity.
For example in H-Cl
formal charge H=0, Cl=0 oxidation number H=1 Cl=-1
Mg2+ is isoelectronic with Ne (period 3). However, P is in period 4 and therefore its electrons exhibit greater shielding than the electrons in Mg2+. Therefore, P's atomic radius is larger.
and since its 2+, it lost two electrons. so the electron repulsions within the atom will be greatly reduced and its 12 protons will pull the 10 electrons easier and with greater attraction
10 comments:
i calculated B
haha v just went over this today in our class ... i got A
The formal charge of boron in BF4- is +3(D).Is not to calculated, is simple B +3 and F-1*4 is BF4-.
The right answer is A.
Boron has three valence electrons in his outer shell and there are 4 electrons around him in BF4-.
So 3-4=-1
+3 is the OXIDATION NUMBER which is differently from formal charge.
whats the difference between a formal charge and an oxidation number?
whats the difference between a formal charge and an oxidation charge?
Formal charge is defined as the difference between the number of valence electrons and the number of Lewis electrons,i.e all unshared electrons on the atom plus the half of all the electrons shared by the atom in the Lewis structure.
Oxidation number is defined as the difference between the number of valence electrons and the number of electrons the atom has in the compound. The number of electrons the atom has assigned in the compound includes all unshared electrons plus half of the electrons shared with atoms of equal electronegativity plus all electrons shared with atoms of lesser electronegativity.
For example in H-Cl
formal charge H=0, Cl=0
oxidation number H=1 Cl=-1
question:
why is Mg2+ smaller in atomic size than P?
Thanks again
Mg2+ is isoelectronic with Ne (period 3). However, P is in period 4 and therefore its electrons exhibit greater shielding than the electrons in Mg2+. Therefore, P's atomic radius is larger.
and since its 2+, it lost two electrons. so the electron repulsions within the atom will be greatly reduced and its 12 protons will pull the 10 electrons easier and with greater attraction
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