Buffers - mixtures of conjugate acid and conjugate base at ±1 pH unit from pH = pKa. Resistant to changes in pH in response to small additions of H+ or OH-. ... Polyprotic acids - dissociation of each H+ can be treated separately if the pKa values are different
They react to form salt ZnCl + hydrogen gas
Answer:
If the electronegativity difference between bonded atoms are too much high ionic bonds are formed if the electronegativity diference is 0.4 or less than 0.4 non polar covalnet bond formed the difference greater than 0.4 polar covalent bond formed.
Explanation:
Ionic bond:
It is the bond which is formed by the transfer of electron from one atom to the atom of another element.
Both bonded atoms have very large electronegativity difference. The atom with large electronegativity value accept the electron from other with smaller value of electronegativity.
For example:
Sodium chloride is ionic compound. The electronegativity of chlorine is 3.16 and for sodium is 0.93. There is large difference is present. That's why electron from sodium is transfer to the chlorine. Sodium becomes positive and chlorine becomes negative ion.
Covalent bond:
It is formed by the sharing of electron pair between bonded atoms.
The atom with larger electronegativity attract the electron pair more towards it self and becomes partial negative while the other atom becomes partial positive.
For example:
In water the electronegativity of oxygen is 3.44 and hydrogen is 2.2. That's why electron pair attracted more towards oxygen, thus oxygen becomes partial negative and hydrogen becomes partial positive.
Atomic Number of Lithium is 3, so it has 3 electrons in its neutral state. Also, Li₂ will have 6 electrons. But the chemical formula we are given has a negative charge on it (i.e Li₂⁻) so there is an additional electron (RED) present on this compound. So, the total number of electrons are 7. The
MOT diagram for this compound is shown below. According to diagram we are having 4 electrons in Bonding Molecular Orbitals (
BMO) and 3 electrons in Anti-Bonding Molecular Orbitals (
ABMO). Bond Order is calculated as,
Bond Order = (# of e⁻s in BMO - # of e⁻s in ABMO) ÷ 2
Bond Order = (4 - 3) ÷ 2
Bond Order = 1 ÷ 2
Or,
Bond Order = 1/2Or,
Bond Order = 0.5
