What is a bond between a positive and negative ion called?
2 answers:
Answer: Option (b) is the correct answer.
Explanation:
An ionic bond is defined as the bond formed due to transfer of electron(s) from one atom to another.
An ionic bond will always be formed between a metal and a non-metal.
For example, sodium metal on donating its one valence electron to chlorine atom will lead to the formation of NaCl compound.
This is because an atom which loses its valence electrons (metals) acquires a positive charge and another atom which gains the electrons (non-metals) acquires a negative charge.
Hence, these opposite charges strongly gets attracted towards each other forming a strong bond.
Whereas in a covalent bond, there will be sharing of electrons between the combining atoms.
For example, an molecule is formed due to sharing of electrons between the two oxygen atoms.
Thus, we can conclude that a bond between a positive and negative ion called an ionic bond.
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Answer:
An example of oxygen–hemoglobin (O2–Hb) dissociation curves from (A) one penguin at pH 7.5, 7.4 and 7.3, and (B) the emperor penguin, the bar-headed goose (Anser indicus) (Black and Tenney, 1980) and the domestic duck (Anas platyrhynchos, forma domestica) (Hudson and Jones, 1986) at pH 7.4. Note that as for the bar-headed goose, the O2–Hb dissociation curve of the emperor penguin is significantly left-shifted as compared with the domestic duck (and most birds). The bar-headed goose photo is courtesy of Graham Scott; the domestic duck photo is by Maren Winter (licensed under the terms of the GNU Free Documentation License, Version 1.2 or any later version); the penguin photo is by J.M.
Explanation:
The resulting regression equations from the plots of log[SO2/(100–SO2)] vs log(PO2) (all saturation points, all penguins combined) were:
pH 7.5: log[SO2/(100–SO2)] = 2.92589 × log(PO2) – 4.24338 (N=43, r2=0.98, P<0.0001),
pH 7.4: log[SO2/(100–SO2)] = 2.94767 × log(PO2) – 4.39858 (N=70, r2=0.98, P<0.0001),
pH 7.3: log[SO2/(100–SO2)] = 3.04945 × log(PO2) – 4.72019 (N=38, r2=0.99, P<0.0001),
pH 7.2: log[SO2/(100–SO2)] = 3.15958 × log(PO2) – 4.97618 (N=9, r2=0.99, P<0.0001).
