The ions of NaF and MgO have the same number of electrons and inter nuclear distances are about the same (235 pm and 215 pm). Why are then the melting points of NaF and MgO so different `(992^(@)C and 2642^(@)C)`?
The ions of NaF and MgO have the same number of electrons and inter nuclear distances are about the same (235 pm and 215 pm). Why are then the melting points of NaF and MgO so different `(992^(@)C and 2642^(@)C)`?
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To understand why the melting points of NaF (sodium fluoride) and MgO (magnesium oxide) are so different despite having the same number of electrons and similar inter-nuclear distances, we can analyze the factors affecting their melting points step by step.
### Step-by-Step Solution:
1. **Identify the Ions and Their Charges:**
- Sodium fluoride (NaF) consists of Na⁺ and F⁻ ions.
- Magnesium oxide (MgO) consists of Mg²⁺ and O²⁻ ions.
...
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Why the melting points of NaF and MgO (990^(@)C and 2640^(@)C ) are so different although both have same number of electrons and nearly the same internuclear distance (236 pm and 216 pm)?
It is not possible to measure the atomic radius precisely since the electron cloud surrounding the atom does not have a sharp boundary. One practical approach to estimate the size of an atom of a non-metallic element is to measure the distance between two atoms when they are bound together by a single bond in a covalent molecule and then dividing by two. For metals we define the term "metallic radius" which is taken as half the inter nuclear distance separating the metal cores in the metallic crystal. Then van der waal's radius represents the over all size of the atoms which includes its valence shell in a non bonded situation. It is the half of the distance between two similar atoms in separate molecules in a solid. The atomic radius decreases across a period and increases down the group. Same trends are observed in case of ionic radius. Ionic radius of the species having same number of electrons depends on the number of protons in their nuclei. Sometimes, atomic and ionic radii give unexpected trends due to poor shielding of nuclear charge by d -and f- orbital electrons. Now answer the following three questions: Which of the following relations is correct, if considered for the same element.
It is not possible to measure the atomic radius precisely since the electron cloud surrounding the atom does not have a sharp boundary. One practical approach to estimate the size of an atom of a non-metallic element is to measure the distance between two atoms when they are bound together by a single bond in a covalent molecule and then dividing by two. For metals we define the term "metallic radius" which is taken as half the inter nuclear distance separating the metal cores in the metallic crystal. Then van der waal's radius represents the over all size of the atoms which includes its valence shell in a non bonded situation. It is the half of the distance between two similar atoms in separate molecules in a solid. The atomic radius decreases across a period and increases down the group. Same trends are observed in case of ionic radius. Ionic radius of the species having same number of electrons depends on the number of protons in their nuclei. Sometimes, atomic and ionic radii give unexpected trends due to poor shielding of nuclear charge by d -and f- orbital electrons. Now answer the following three questions: K^(+), CI^(-), Ca^(2+), S^(2-) ions are isoelectronic. The decreasing order of their size is:
D (+) Glucose has melting point 146^@ C and specific rotation [alpha]_(D)^(25) is + 112^@ C. Another D(+) Glucose has melting point 150^@ C and specific rotational [alpha]_(D)^(25) is + 18.7^@ C. The two form have significantly different optical rotation but when an aqueous solution of either form is allowed to stand, it rotation changes. The specific rotation of one form decreases and rotation of other increases until both solution show the same value +52.7^@ . The change in rotation towards an equilibrium value is called mutarotation. Mutarotation is characteristic feature of:
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D (+) Glucose has melting point 146^@ C and specific rotation [alpha]_(D)^(25) is + 112^@ C. Another D(+) Glucose has melting point 150^@ C and specific rotational [alpha]_(D)^(25) is + 18.7^@ C. The two form have significantly different optical rotation but when an aqueous solution of either form is allowed to stand, it rotation changes. The specific rotation of one form decreases and rotation of other increases until both solution show the same value +52.7^@ . The change in rotation towards an equilibrium value is called mutarotation. For mannose the mutarotation can be shown in brief as follow:
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